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Industrial Chemistry Process - Practice Questions (34)

Question 1: 1. The development of colloid chemistry has led to the widespread use of nanomaterials. The analysis...

1. The development of colloid chemistry has led to the widespread use of nanomaterials. The analysis and application of the process below does not make sense $\_\_\_\_$ $\mathrm { Fe } \xrightarrow [ \text { I } ] { \mathrm { Cl } _ { 2 } } \mathrm { FeCl } \xrightarrow [ \text { II } ] { \text { 试剂(1)} } \mathrm { Fe } ( \mathrm { OH } ) _ { 3 }$ colloid

  • A. A. Iron can react with $\mathrm { Cl } _ { 2 }$ and liquid chlorine cannot be stored in dry cylinders.
  • B. B. Reagent (1) is boiling water, which produces a reddish-brown liquid
  • C. C. Iron hydroxide colloids with dispersant particle diameters in the $1 \sim 100 \mathrm {~nm}$ range
  • D. D. The properties of a material can be changed by changing its particle size.

Answer: A

Solution: A. Dry $\mathrm { Cl } _ { 2 }$ at room temperature does not react with iron, and liquid chlorine can be stored in dry cylinders; B. $\mathrm { FeCl } _ { 3 }$ hydrolyzes in boiling water to form $\mathrm { Fe } ( \mathrm { OH } ) _ { 3 }$ colloids, and $\mathrm { Fe } ( \mathrm { OH } ) _ { 3 }$ colloids are reddish-brown; C. The diameter of colloidal particles ranges from $1 \sim 100 \mathrm {~nm}$, C is correct; D. The properties of nanomaterials are related to the particle size, and changing the particle size can regulate the properties, D is correct;

Question 2: 2. Cobalt and its compounds have a wide range of applications in industry, electronics and other fie...

2. Cobalt and its compounds have a wide range of applications in industry, electronics and other fields. A partial process for extracting cobalt from zinc smelting waste residue (mainly containing + 2-valent oxides or monomers of cobalt, zinc, lead, iron and manganese) is as follows: ![](/images/questions/industrial-chemistry/image-001.jpg) It is known that: $\mathrm { K } _ { \mathrm { sp } } \left[ \mathrm { Fe } ( \mathrm { OH } ) _ { 3 } \right] = 1.0 \times 10 ^ { - 39 } , \mathrm {~K} _ { \mathrm { sp } } \left[ \mathrm { Co } ( \mathrm { OH } ) _ { 3 } \right] = 1.0 \times 10 ^ { - 44 } , \mathrm { Zn } ( \mathrm { OH } ) _ { 2 }$ and $\mathrm { Al } ( \mathrm { OH } ) _ { 3 \text { 的化 } }$ have similar chemical properties. The following statements are correct $$ \mathrm { ClO } ^ { - } + 2 \mathrm { Co } ^ { 2 + } + \mathrm { H } _ { 2 } \mathrm { O } + 4 \mathrm { OH } ^ { - } = \mathrm { Cl } ^ { - } + 2 \mathrm { Co } ( \mathrm { OH } ) _ { 3 } \downarrow $$

  • A. A. It is known that the $\mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 8 }$ has ${ } ^ { S }$ elements with ${ } ^ { + 6 } , 1 \mathrm { molNa } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 8 }$ containing ${ } ^ { 2 \mathrm {~N} _ { \mathrm { A } } }$ peroxide bonds
  • B. B. Combined with process analysis, $\mathrm { Fe } ^ { 2 + }$ is more reductive than $\mathrm { Co } ^ { 2 + }$
  • C. C. Use FeS to remove $\mathrm { Zn } ^ { 2 + } , \mathrm { pH }$ from the filtrate. The larger the $\mathrm { Zn } ^ { 2 + }$ is, the better it is for removing $\mathrm { Zn } ^ { 2 + }$ from the filtrate.
  • D. D. Ionic equation for "oxidized cobalt

Answer: B

Solution: A. The highest valence of the element S is $+ 6 , ~ \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 8 }$, the element ${ } ^ { \mathrm { S } }$ has the valence ${ } ^ { + 6 }$, so there is a -1-valent oxygen atom in $\mathrm { S } _ { 2 } \mathrm { O } _ { 8 } ^ { 2 - }$, and there is a -2-valent oxygen atom in $\mathrm { S } _ { 2 } \mathrm { O } _ { 8 } ^ { 2 - }$, and there is an oxygen atom in $\mathrm { S } _ { 2 } \mathrm { O } _ { 8 } ^ { 2 - }$. According to its structural formula <img class="imgSvg" id = "mi1n1jemfg0vp8b6m9" src="data:image/svg+xml;base64, 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 +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW4xamVtZmcwdnA4YjZtOS0zIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjIwNS42Nzg4MDEzMTUyMjE4MiIgeTE9IjY0LjAyOTg2NTkxNjg4MjY3IiB4Mj0iMjMyLjk1ODYxMjEzNzQwNTg1IiB5Mj0iNDguMjc5ODg0MjgxNzg2OTkiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW4xamVtZmcwdnA4YjZtOS01IiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjE4Ny40NzM2MDE4ODQ4NjQ1NiIgeTE9Ijg5Ljg5MjE1Mzg4MDI2NTE3IiB4Mj0iMjAzLjIyMzYyMDI0OTc2MTcyIiB5Mj0iNjIuNjEyMzY0MjY0MDQxOTMiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW4xamVtZmcwdnA4YjZtOS03IiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjE5Mi4zODM5NjQwMTU3ODQ3NiIgeTE9IjkyLjcyNzE1NzE4NTk0NjY1IiB4Mj0iMjA4LjEzMzk4MjM4MDY4MTIHkyPSI2NS40NDczNjc1Njk3MjM0MSI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbjFqZW1mZzB2cDhiNm05LTkiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMjAzLjIyMzYxODM0MTIyNTI2IiB5MT0iNjUuNDQ3MzY0MjY0MDQxMjgiIHgyPSIyMTguOTczNTk5OTc2MzIwOTQiIHkyPSI5Mi43MjcxNzUwODYyMjUyOSI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbjFqZW1mZzB2cDhiNm05LTExIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjIwOC4xMzM5ODQyODkyMTgzOCIgeTE9IjYyLjYxMjM2NzU2OTcyNDA2IiB4Mj0iMjIzLjg4Mzk2NTkyNDMxNDA1IiB5Mj0iODkuODkyMTc4MzkxOTA4MDYiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW4xamVtZmcwdnA4YjZtOS0xMyIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSIxNzguMzk5MDExNjk4OTk4NTgiIHkxPSI0OC4yNzk4NDc1NTE5ODU1MSIgeDI9IjIwNS42Nzg4MDEzMTUyMjE4MiIgeTI9IjY0LjAyOTg2NTkxNjg4MjY3Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFuMWplbWZnMHZwOGI2bTktMTUiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMTUxLjExOTIwMDg3NjgxNDU1IiB5MT0iNjQuMDI5ODI5MTg3MDgxMTkiIHgyPSIxNzguMzk5MDExNjk4OTk4NTgiIHkyPSI0OC4yNzk4NDc1NTE5ODU1MSI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbjFqZW1mZzB2cDhiNm05LTE3IiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjEyMy44Mzk0MTEyNjA1OTEyOSIgeTE9IjQ4LjI3OTgxMDgyMjE4NDAzIiB4Mj0iMTUxLjExOTIwMDg3NjgxNDU1IiB5Mj0iNjQuMDI5ODI5MTg3MDgxMTkiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW4xamVtZmcwdnA4YjZtOS0xOSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSIxMTAuNTQ0NjEyNTk5NDkyMTUiIHkxPSIxOS41ODI1MDE2NTI4NDEzODciIHgyPSIxMjYuMjk0NTk0MjM0NTg3ODUiIHkyPSI0Ni44NjIzMTI0NzUwMjU0MiI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. 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PC90ZXh0Pjx0ZXh0IHg9IjY5LjI3OTgxMDgyMjE4NDAyIiB5PSI0OC4yNzk3NzQwOTIzODI1NSIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgIj48L3RleHQ + PHRleHQgeD0iNDIiIHk9IjY0LjAyOTc1NTcyNzQ3ODIzIiBjbGFzcz0iZGVidWciIGZpbGw9IiNmZjAwMDAiIHN0eWxlPSIKICAgICAgICAgICAgICAgIGZvbnQ6IDVweCBEcm9pZCBTYW5zLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAgICAiPjwvdGV4dD48L2c +PC9zdmc+"/> A. $1 \mathrm { molNa } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 8 } \quad \mathrm {~N} _ { \mathrm { A } }$ contains a peroxy bond, A is wrong; B. From the process, $\mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 8 }$ oxidizes $\mathrm { Fe } ^ { 2 + }$ to $\mathrm { Fe } ^ { 3 + }$ when $\mathrm { Fe } ^ { 3 + }$ is added and $\mathrm { Co } ^ { 2 + }$ is not oxidized. In the subsequent operation, $\mathrm { Co } ^ { 2 + }$ is oxidized with NaClO, so it can be seen that $\mathrm { Fe } ^ { 2 + }$ is more reductive than $\mathrm { Co } ^ { 2 + }$, B is correct; C. $\mathrm { Zn } ( \mathrm { OH } ) _ { 2 }$ and $\mathrm { Al } ( \mathrm { OH } ) _ { 3 }$ are chemically similar and can be dissolved in strong bases. $\mathrm { pH } _ { \text {过高时,} } \mathrm { Zn } ^ { 2 + }$ reacts with $\mathrm { OH } ^ { - }$ to form a soluble substance, which makes it impossible to remove the zinc element, C is wrong; D. According to process analysis, the reactants are $\mathrm { ClO } ^ { - }$ and $\mathrm { Co } ^ { 2 + }$, and the formation of $\mathrm { Cl } ^ { - }$ and $\mathrm { Co } ( \mathrm { OH } ) _ { 3 } , \mathrm { pH } = 5$ belongs to the acidic nature, and according to the conservation of gain and loss of electrons, the conservation of ionic charge, and the conservation of elements to be leveled, the The ionic equation is $\mathrm { ClO } ^ { - } + 2 \mathrm { Co } ^ { 2 + } + 5 \mathrm { H } _ { 2 } \mathrm { O } = \mathrm { Cl } ^ { - } + 2 \mathrm { Co } ( \mathrm { OH } ) _ { 3 } \downarrow + 4 \mathrm { H } ^ { + } \quad , ~ \mathrm { D }$ Wrong;

Question 3: 3. Sodium borohydride is a white powder, commonly used as a reducing agent in organic synthesis, and...

3. Sodium borohydride is a white powder, commonly used as a reducing agent in organic synthesis, and decomposes in moist air. Sodium metaborate ( $\mathrm { NaBO } _ { 2 }$ ${ } ^ { 2 }$) is soluble in water, insoluble in ethanol, easily hydrolyzed, but stable under alkaline conditions. The procedure for making $\mathrm { NaBH } _ { 4 }$ from boron concentrate (which contains a certain amount of $\mathrm { B } _ { 2 } \mathrm { O } _ { 3 }$, and impurities such as $\mathrm { Al } _ { 2 } \mathrm { O } _ { 3 } , \mathrm { SiO } _ { 2 } , \mathrm { Fe } _ { 2 } \mathrm { O } _ { 3 }$), and analyzing the purity of $\mathrm { NaBH } _ { 4 }$ samples (the impurities do not take part in the reaction) is as follows. ![](/images/questions/industrial-chemistry/image-002.jpg) Step I : Take $\mathrm { mg } _ { \text {样品配成 } } 250 \mathrm {~mL} _ { \text {溶液 } }$, take ${ } ^ { 25.00 m L }$ in an iodine measuring flask, and add $\mathrm { V } _ { 1 } \mathrm { ml } \mathrm { c } _ { 1 } \mathrm {~mol} \cdot \mathrm {~L}$ of $\mathrm { KIO } _ { 3 \text { 溶液,发生反应:} } \mathrm { NaBH } _ { 4 } + \mathrm { KIO } _ { 3 } \rightarrow \mathrm { NaBO } _ { 2 } + \mathrm { KI } + \mathrm { H } _ { 2 } \mathrm { O }$ (the chemical equation is not leveled). Step II: Add excess ${ } ^ { \mathrm { KI } }$ solution to the solution after the reaction in Step I, adjust ${ } ^ { \mathrm { pH } }$ to convert $\mathrm { KIO } _ { 3 }$ to ${ } ^ { \mathrm { I } _ { 2 } }$, cool, and leave it in the dark for a few minutes. Step III: Adjust ${ } ^ { \mathrm { pH } }$ of the mixture obtained in Step II to weak acidity, add a few drops of starch solution, titrate with ${ } ^ { \mathrm { C } _ { 2 } \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 } }$ standard solution to the end point, a reaction occurs: $\mathrm { I } _ { 2 } + 2 \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 } = 2 \mathrm { NaI } + \mathrm { Na } _ { 2 } \mathrm {~S} _ { 4 } \mathrm { O } _ { 6 }$, consuming $\mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 }$ standard solution $\mathrm { I } _ { 2 } + 2 \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 } = 2 \mathrm { NaI } + \mathrm { Na } _ { 2 } \mathrm {~S} _ { 4 } \mathrm { O } _ { 6 }$. _17]] of standard solution $\mathrm { V } _ { 2 } \mathrm {~mL}$. The following operations related to this experiment are correct ( )

  • A. A. The reagent used to adjust the pH in "Step II" can be a suitable amount of KOH solution.
  • B. B. Inadvertent adjustment of the mixture to strong acidity during "Step III" will result in a high purity of the measured sample.
  • C. C. When the last half drop of $\mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 }$ standard solution is added and the blue color of the solution disappears, the end point of the titration is reached.
  • D. D. When reading, looking flat before titration and looking down after titration will result in a large purity of the measured sample

Answer: D

Solution: A. $\mathrm { I } _ { 2 }$ can react with KOH solution, so the reagent used to adjust the pH in "Step II" cannot be KOH solution, so A is wrong; B. "Step III" of the operation process inadvertently adjust the mixture to a strong acid, then the reaction will occur [INLINE_FORMULA_1]], consume $\mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 }$, according to the reaction $\mathrm { I } _ { 2 } + 2 \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 } = 2 \mathrm { NaI } + \mathrm { Na } _ { 2 } \mathrm {~S} _ { 4 } \mathrm { O } _ { 6 }$ can be measured $\mathrm { I } _ { 2 }$ of the amount of material is too large, and $\mathrm { I } _ { 2 }$ is $\mathrm { I } _ { 2 }$. INLINE_FORMULA_6]] is converted from $\mathrm { KIO } _ { 3 }$, and the total amount of substance in $\mathrm { KIO } _ { 3 }$ is constant, the measured $\mathrm { I } _ { 2 }$ consumes a large amount of $\mathrm { KIO } _ { 3 }$, which is known to cause the measured amount of $\mathrm { NaBH } _ { 4 } + \mathrm { KIO } _ { 3 } \rightarrow \mathrm { NaBO } _ { 2 } + \mathrm { KI } + \mathrm { H } _ { 2 } \mathrm { O }$ to increase. According to the reaction $\mathrm { NaBH } _ { 4 } + \mathrm { KIO } _ { 3 } \rightarrow \mathrm { NaBO } _ { 2 } + \mathrm { KI } + \mathrm { H } _ { 2 } \mathrm { O }$, the amount of $\mathrm { KIO } _ { 3 }$ consumed by $\mathrm { KIO } _ { 3 }$ is too large, which can be known from the reaction $\mathrm { KIO } _ { 3 }$, and it will result in the amount of $\mathrm { KIO } _ { 3 }$ measured by the reaction $\mathrm { KIO } _ { 3 }$ being small, that is to say, the purity of the sample is small, so B is incorrect. C. The last half drop of $\mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 }$ standard solution is added, the blue color of the solution disappears and does not return to its original color within half a minute to prove that the end point of titration is reached; D. When taking a reading, looking down before titration will lead to a small reading, ${ } ^ { \mathrm { V } _ { 2 } }$ is small, and according to the reaction $\mathrm { I } _ { 2 } + 2 \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 } = 2 \mathrm { NaI } + \mathrm { Na } _ { 2 } \mathrm {~S} _ { 4 } \mathrm { O } _ { 6 }$, the measured quantity of $\mathrm { I } _ { 2 }$ is small, and $\mathrm { I } _ { 2 }$ is $\mathrm { I } _ { 2 }$, the quantity of $\mathrm { I } _ { 2 }$ is $\mathrm { I } _ { 2 }$. ]] is converted from $\mathrm { KIO } _ { 3 }$, and the total amount of substance in $\mathrm { KIO } _ { 3 }$ is constant, the measured amount of $\mathrm { I } _ { 2 }$ consumed by $\mathrm { KIO } _ { 3 }$ is small, and the amount of $\mathrm { KIO } _ { 3 }$ consumed by $\mathrm { KIO } _ { 3 }$ is small, and the $\mathrm { KIO } _ { 3 }$ is small. FORMULA_22]] is constant, the amount of $\mathrm { NaBH } _ { 4 } + \mathrm { KIO } _ { 3 } \rightarrow \mathrm { NaBO } _ { 2 } + \mathrm { KI } + \mathrm { H } _ { 2 } \mathrm { O } _ { \text {可知,会导致测得的and } } \mathrm { NaBH } _ { 4 }$ consumed $\mathrm { KIO } _ { 3 }$ is large, which means that the purity of the sample is large, so D is correct; Therefore, the answer is D.

Question 4: 4. Known: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid solution i...

4. Known: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid solution is colorless, ferric ellagic acid solution is blue-black, according to the analysis of the experimental process shown in the figure, the following narrative is incorrect () green tea leaves in water $- \underset { \text {(1)} } { - \underset { \text { add } } { \mathrm { FeSO } _ { 4 } } } \rightarrow \underset { \text { 蓝黑色 } } { - \underset { \text {(2)} } { \text { 维生入素 } } \underset { \text { 加 } } { \mathrm { C } } } \longrightarrow \underset { \text { 复为原色 } } { - \underset { \text {(3)} } { \text { drop in } } } \rightarrow \underset { \text { 不变色 } } { \text { KSCN solution } } - \underset { \text {(4)drop in } } { \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } }$ solution $\rightarrow$ turned red

  • A. A. The ferrous ellagic acid produced by the reaction of green tea leaf water with ${ } ^ { \mathrm { FeSO } _ { 4 } }$ is easily oxidized to ferric ellagic acid, which is blue-black in color.
  • B. B. Vitamin C reduces ferric ellagic acid to ferrous ellagic acid
  • C. C. ${ } ^ { \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } }$ oxidizes $\mathrm { Fe } ^ { 2 + }$ to $\mathrm { Fe } ^ { 3 + }$ in (4).
  • D. D. When taking iron supplements, it's best to drink some green tea water

Answer: D

Solution: A. Ferrous ellagic acid solution is colorless, ferric ellagic acid solution is blue-black, add $\mathrm { FeSO } _ { 4 }$ to the green tea water, blue-black, indicating that ferrous ellagic acid can be easily oxidized by oxygen, A is correct; B. To the blue-black solution add vitamin C, the solution from blue-black color back to the original color, indicating that vitamin C can be iron ellagic acid also B. When vitamin C is added to a blue-black solution, the solution returns to its original color from blue-black, indicating that vitamin C can reduce ferric ellagic acid to ferrous ellagic acid, which is a reducing property; C. The solution eventually turns red in process (4), indicating that $\mathrm { Fe } ^ { 3 + }$ is generated, and $\mathrm { Fe } ^ { 2 + }$ is oxidized by $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ to produce $\mathrm { Fe } ^ { 3 + } , \mathrm { Fe } ^ { 3 + }$ and $\mathrm { SCN } ^ { - }$. complexation reaction occurs to make the solution red, C is correct; D. Because the ellagic acid in the water of green tea leaves reacts with $\mathrm { Fe } ^ { 2 + }$ in the iron supplements to produce ferrous ellagic acid, which is unstable and easily oxidized to ferrous ellagic acid, you can't drink green tea when you take iron supplements, D. Wrong; Therefore, this question is D.

Question 5: 5. A qualitative test of some ions in a sodium salt solution is performed as follows. The following ...

5. A qualitative test of some ions in a sodium salt solution is performed as follows. The following analysis is correct ![](/images/questions/industrial-chemistry/image-003.jpg) Known: ${ } ^ { - }$ in solution reacts with $\mathrm { Cu } ^ { 2 + }$ to form ${ } ^ { \mathrm { I } _ { 2 } }$ and $\mathrm { CuI } ^ { - }$ white precipitates.

  • A. A. Step (2) operation is filtration and step (3) operation is distillation
  • B. B. The ionic equation for the reaction of solution Y with hydrogen peroxide is $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 } + 2 \mathrm { I } ^ { - } = \mathrm { I } _ { 2 } + 2 \mathrm { OH }$
  • C. C. To test for the presence of $\mathrm { I } ^ { - }$ in this sodium salt solution, add silver nitrate solution acidified with nitric acid directly to the sodium salt solution
  • D. D. The anions in this sodium salt solution react with $\mathrm { Cu } ^ { 2 + }$ in both complex decomposition and redox reactions

Answer: D

Solution: A. Step (2) operation for the separation of solid-liquid operation, is filtration; step (3) operation for the separation of organic layer, aqueous layer of the operation, for the extraction and separation, A error; B. Solution Y reacts with hydrogen peroxide and iodide ions to produce electrons and water, $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 } + 2 \mathrm { I } ^ { + } + 2 \mathrm { H } ^ { + } = \mathrm { I } _ { 2 } + 2 \mathrm { H } _ { 2 } \mathrm { O }$, B error; C. Add nitric acid acidified silver nitrate solution, nitric acid will oxidize iodide ions and sulfur ions to sulfur monomers and iodine monomers, C error; D. It is known that: ${ } ^ { I }$ in solution will react with $\mathrm { Cu } ^ { 2 + }$ to form ${ } ^ { I }$ and $\mathrm { CuI } ^ { - }$ white precipitates; sulfur ions in the solution of this sodium salt and [[INLINE_FORMULA_5] ] can react with $\mathrm { Cu } ^ { 2 + }$ to produce copper sulfide precipitate, and iodine ions can react with copper ions to produce iodine monomers and CuI, D is correct;

Question 6: 6. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid...

6. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid solution is colorless, ferric ellagic acid solution is blue-black. According to the analysis of the experimental process shown in the figure, the following narrative is wrong Green tea leaves water $\underset { \text {(1)} \text { add } } { \mathrm { FeSO } _ { 4 } }$ blue-black $\underset { \text {(2)add } } { \text { 维生素 } \mathrm { C } }$ return to the original color $\underset { \text {(3)drop in } } { \mathrm { KSCN } _ { \text {溶液 } } }$ does not change color $\underset { \text {(4)drop in } } { \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } \text { 溶液 } }$ change to red

  • A. A. The ferrous ellagic acid produced by the reaction of green tea leaf water with $\mathrm { FeSO } _ { 4 }$ is easily oxidized to ferric ellagic acid, which is blue-black in color.
  • B. B. Vitamin C reduces ferric ellagic acid to ferrous ellagic acid
  • C. C. (2 reactions occur in (4)
  • D. D. You should not drink green tea when taking iron supplements

Answer: C

Solution: A. Ferrous ellagic acid solution is colorless and ferric ellagic acid solution is blue-black. Adding $\mathrm { FeSO } _ { 4 }$ to the water of green tea leaves results in a blue-black color, which indicates that ferrous ellagic acid can be easily oxidized by oxygen, A is correct; B. Add vitamin C to the blue-black solution, and the solution returns to its original color from blue-black, indicating that vitamin C can reduce the ferric ellagic acid to ferrous ellagic acid, which shows the reducing property, B is correct; C. The reactions in (4) include the oxidation of vitamin $\mathrm { C } , \mathrm { Fe } ^ { 2 + }$ by $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ and the complexation reaction between $\mathrm { Fe } ^ { 3 + }$ and SCN; D. Because the ellagic acid in green tea leaf water reacts with $\mathrm { Fe } ^ { 2 + }$ in iron supplements to produce ferrous ellagate, which is unstable and easily oxidized to ferrous ellagate, so you can't drink green tea when you take iron supplements, D. Correct; Therefore, the answer is C.

Question 7: 7. The main process of preparing anhydrous $\mathrm { NaHSO } _ { 3 }$ from ${ } ^ { \mathrm { SO } ...

7. The main process of preparing anhydrous $\mathrm { NaHSO } _ { 3 }$ from ${ } ^ { \mathrm { SO } _ { 2 } }$ and soda ash as raw materials in industry is shown in the figure. The following analysis is correct ( ) ![](/images/questions/industrial-chemistry/image-004.jpg)

  • A. A. Masterbatch of $\mathrm { pH } > 7$
  • B. B. The neutralized solution has more than two solutes in it
  • C. C. Only glass rods and funnels are needed to simulate the separation of wet materials and mother liquors in the laboratory.
  • D. D. Magnetite is produced when $\mathrm { SO } _ { 2 }$ is industrially prepared by reacting pyrite with oxygen at high temperatures

Answer: B

Solution: A. The mother liquor is the solution after crystallization $\mathrm { NaHSO } _ { 3 }$, $\mathrm { NaHSO } _ { 3 }$ ionization is more acidic than hydrolysis in the $\mathrm { HSO } _ { 3 } ^ { - }$ solution ($\mathrm { pH } < 7$), and the mother liquor needs to be neutralized to $\mathrm { pH } = 8$, indicating that the mother liquor itself $\mathrm { pH } < 7 , \mathrm {~A}$ is wrong. _4]], indicating that the mother liquor itself $\mathrm { pH } < 7 , \mathrm {~A}$ is incorrect; B. In the neutralization step, the mother liquor (containing $\mathrm { NaHSO } _ { 3 }$) reacts with soda ash ($\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$), and $\mathrm { HSO } _ { 3 } ^ { - }$ reacts with $\mathrm { CO } _ { 3 } ^ { 2 - }$ to form $\mathrm { SO } _ { 3 } ^ { 2 - }$ and [$\mathrm { CO } _ { 3 } ^ { 2 - }$. INLINE_FORMULA_11]] in an alkaline solution may have $\mathrm { Na } _ { 2 } \mathrm { SO } _ { 3 } , \mathrm { NaHCO } _ { 3 }$ and an excess of $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$, which is more than two types of solutes, B is correct; C. Separating the wet material (crystals) from the mother liquor is a filtration operation, which requires a funnel, a glass rod, a beaker, and filter paper, etc., and using only a glass rod and a funnel is not enough, C is wrong; D. Pyrite ($\mathrm { FeS } _ { 2 }$) reacts with oxygen at high temperature to form ${ } ^ { \mathrm { SO } _ { 2 } }$ and $\mathrm { Fe } _ { 2 } \mathrm { O } _ { 3 }$ (hematite), not magnetite ($\mathrm { Fe } _ { 3 } \mathrm { O } _ { 4 }$). D is wrong ; Therefore, the answer is B.

Question 8: 9. Ethylene oxide (b, molecular formula for $\mathrm { C } _ { 2 } \mathrm { H } _ { 4 } \mathrm { O...

9. Ethylene oxide (b, molecular formula for $\mathrm { C } _ { 2 } \mathrm { H } _ { 4 } \mathrm { O }$, relative molecular mass of 44) is a colorless liquid at room temperature, is an important industrial synthesis intermediates and disinfectants. Analyzing the flow shown in the figure, the following statement is not correct ( ) ![](/images/questions/industrial-chemistry/image-005.jpg)

  • A. A. b, c are both more soluble in water
  • B. B. $a$, $b$ Molecule with all atoms coplanar
  • C. C. b and c both denature proteins
  • D. D. a The addition products of $\mathrm { H } _ { 2 } \mathrm { O }$ and c are alcohols.

Answer: B

Solution: A. b and carbon dioxide have the same relative molecular mass, and carbon dioxide is a gas, and ethylene oxide is a liquid at low temperatures, indicating that ethylene oxide is a polar molecule, soluble in water, and c can form hydrogen bonds with water molecules and is easily soluble in water, so A is correct; B. b contains saturated carbon atoms and has the structural characteristics of methane, then all atoms in the molecule are not coplanar, so B is wrong; C. b can be used as a disinfectant and can denature proteins. c is similar in nature to ethanol and can denature proteins, so C is correct; D. a and water addition reaction to generate ethanol, belonging to the alcohol, so D correct.

Question 9: 10. The laboratory process for simulating the removal of ammonia nitrogen ($\mathrm { NH } _ { 3 }$)...

10. The laboratory process for simulating the removal of ammonia nitrogen ($\mathrm { NH } _ { 3 }$) from wastewater by an industrial bleach solution (with NaClO as the active ingredient) is as follows: ![](/images/questions/industrial-chemistry/image-006.jpg) The following analysis is correct

  • A. A. (Refining of crude brine by distillation in (1)
  • B. B. (The main electrode reaction at the anode in (2): $4 \mathrm { OH } ^ { - } - 4 \mathrm { e } ^ { - } = 2 \mathrm { H } _ { 2 } \mathrm { O } + \mathrm { O } _ { 2 } \uparrow$
  • C. C. (Reaction to prepare bleach solution in (3): $\mathrm { Cl } _ { 2 } + \mathrm { OH } ^ { - } = \mathrm { Cl } ^ { - } + \mathrm { HClO }$
  • D. D. (Redox reactions occur in (2), (3) and (4)

Answer: D

Solution: A. (1) in the use of distillation can only get distilled water, do not get refined brine, to get refined brine need to use chemical methods to remove impurities, error; B. The anode undergoes an oxidation reaction, and the main electrode reaction of the anode in (2): $2 \mathrm { Cl } ^ { - } - 2 \mathrm { e } ^ { - } = \mathrm { Cl } _ { 2 } \uparrow$ , is wrong; C. The chlorine gas generated reacts with sodium hydroxide to form sodium hypochlorite, the reaction for preparing bleach solution in (3): $\mathrm { Cl } _ { 2 } + 2 \mathrm { OH } ^ { - } = \mathrm { Cl } ^ { - } + \mathrm { ClO } ^ { - } + \mathrm { H } _ { 2 } \mathrm { O }$ , error ; D. The electrolysis of sodium chloride solution in (2) generates hydrogen and chlorine gas as well as sodium hydroxide, the reaction of chlorine gas with sodium hydroxide in (3) generates sodium chloride and sodium hypochlorite, and the reaction of sodium hypochlorite with ammonia gas in (4) generates nitrogen and sodium chloride, all of which involve changes in elemental valence and redox reactions, correct;

Question 10: 11. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic aci...

11. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid solution is colorless, ferric ellagic acid solution is blue-black. According to the analysis of the experimental process shown in the figure, the following description is wrong ( ) $\underset { \text { 叶水 } } { \text { 绿茶 } } \underset { ( 1 ) \text { add } } { \mathrm { FeSO } _ { 4 } } \underset { \text { 蓝黑色 } } { \text { 维生素 } } \underset { \text {(2)add } } { \text { 恢复为 } } \underset { \text { 原色 } } { \mathrm { KSCN } \text { 溶液 } } \underset { \text {(3)drop in } } { \text { 不变色 } } \xrightarrow [ \text {(4)drop in } ] { \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } \text { 溶液 } } \underset { \text { 红色 } } { \text { 变为 } }$

  • A. A. The ferrous tannate produced by the reaction between green tea leaf water and ferrous sulfate is easily oxidized to ferric tannate and becomes blue-black in color.
  • B. B. Vitamin C in reaction (2) exhibits oxidizing properties
  • C. C. In reaction (4) $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ can be replaced with freshly made chlorine water.
  • D. D. Green tea leaf water should not be taken at the same time as iron supplements

Answer: B

Solution: A. Green tea contains a large amount of ellagic acid and $\mathrm { FeSO } _ { 4 }$ react to produce ferrous ellagic acid, the solution is colorless, ferrous ellagic acid is oxidized to produce ferrous ellagic acid and blue-black, A correct; B. Add vitamin C to the blue-black solution, and the solution returns to its original color from blue-black, which means that vitamin C can reduce ferric ellagic acid to ferrous ellagic acid, which is a reductive property, and B is wrong; C. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ and freshly made chlorine water are both oxidizing, and can oxidize $\mathrm { Fe } ^ { 2 + }$ to $\mathrm { Fe } ^ { 3 + }$, so $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ can be replaced by freshly made chlorine water in reaction (4), C is correct; D. Because the ellagic acid in green tea water and iron supplements in the $\mathrm { Fe } ^ { 2 + }$ reaction to generate ferrous ellagic acid, ferrous ellagic acid is unstable, easy to be oxidized to ferric ellagic acid, so you can not drink green tea when taking iron supplements, D is correct;

Question 11: 12. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic aci...

12. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid solution is colorless, ferric ellagic acid solution is blue-black, according to the analysis of the experimental process as shown in the figure, the following description is not correct $\underset { \text { 叶水 } } { \text { 绿茶 } } \underset { \text {(1)add } } { \mathrm { FeSO } _ { 4 } } \xrightarrow [ \text { 蓝黑色 } ] { \text { Vitamin C } } \xrightarrow [ \text {(2)add } ] { \text { 恢复为 } } \xrightarrow [ \text {(3)drop in } ] { \mathrm { KSCN } \text { 溶液 } }$ does not change color $\xrightarrow [ \text {(4)drop in } ] { \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } \text { 溶液 } }$ becomes

  • A. A. The ferrous ellagic acid produced by the reaction of green tea leaf water with ${ } ^ { \mathrm { FeSO } _ { 4 } }$ is easily oxidized to ferric ellagic acid, which is blue-black in color.
  • B. B. Vitamin C reduces iron ellagic acid to ferrous ellagic acid
  • C. C. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ oxidizes $\mathrm { Fe } ^ { 2 + }$ to $\mathrm { Fe } ^ { 3 + }$ in (4).
  • D. D. When taking iron supplements, it's best to drink some green tea water

Answer: D

Solution: A. Ferrous ellagic acid solution is colorless and ferric ellagic acid solution is blue-black. Adding $\mathrm { FeSO } _ { 4 }$ to green tea leaf water results in a blue-black color, indicating that ferrous ellagic acid is easily oxidized, A is correct; B. Add vitamin C to the blue-black solution, and the solution returns to its original color from blue-black, indicating that vitamin C can reduce iron ellagic acid to ferrous ellagic acid, which is a reductive property, B is correct; C. KSCN solution is added to (3), the solution does not change color, which means that the solution only contains $\mathrm { Fe } ^ { 2 + }$ and does not contain $\mathrm { Fe } ^ { 3 + }$; (4) $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ is added to (4), and the solution turns red, which means that $\mathrm { Fe } ^ { 2 + }$ is replaced by $\mathrm { Fe } ^ { 2 + }$ in (4), which means that $\mathrm { Fe } ^ { 2 + }$ is replaced by $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$. 4]] is oxidized by $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ to $\mathrm { Fe } ^ { 3 + }$ in (4), C is correct; D. Because the ellagic acid in green tea leaf water reacts with $\mathrm { Fe } ^ { 2 + }$ in iron supplements to produce ferrous ellagate, which is unstable and easily oxidized to ferrous ellagate, green tea should not be drunk when taking iron supplements; Therefore, the answer is D.

Question 12: 13. Known: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid solution ...

13. Known: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid solution is colorless, ferric ellagic acid solution is blue-black, generalized as shown in the figure of the experimental process analysis, the following description is wrong $\underset { \text { 叶水 } } { \text { 绿茶 } } \underset { \text {(1)add } } { \mathrm { FeSO } _ { 4 } } \underset { \text { 蓝黑色 } } { \mathrm { FeSO } _ { \text {(2)add } } ^ { \text {维生素 } } \mathrm { C } }$ is restored to $\underset { \text {(3)drop in } } { \mathrm { KSCN } \text { 溶液 } }$ does not change color $\xrightarrow [ \text {(4)drop in } ] { \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } \text { 溶液 } } \underset { \text { 红色 } } { \text { 变为 } }$

  • A. A. The ferrous ellagic acid produced by the reaction of green tea leaf water with $\mathrm { FeSO } _ { 4 }$ is easily oxidized to ferric ellagic acid, which is blue-black in color.
  • B. B. Vitamin C reduces ferric ellagic acid to ferrous ellagic acid
  • C. C. (Only 1 reaction occurs in (4)
  • D. D. You should not drink green tea when taking iron supplements

Answer: C

Solution: A. Ferrous ellagic acid solution is colorless and ferric ellagic acid solution is blue-black. When $\mathrm { FeSO } _ { 4 }$ is added to green tea leaf water, a blue-black color appears, indicating that ferrous ellagic acid is easily oxidized, A is correct; B. Add vitamin C to the blue-black solution, and the solution returns to its original color from blue-black, indicating that vitamin C can reduce the iron ellagic acid to ferrous ellagic acid, which is a reductive property, B is correct; C. The reactions in (4) include vitamin C, oxidation of $\mathrm { Fe } ^ { 2 + }$ by $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$, and complexation of $\mathrm { Fe } ^ { 3 + }$ with $\mathrm { SCN } ^ { - }$, which are more than 2 reactions, C is wrong; D. Because the ellagic acid in green tea leaf water reacts with $\mathrm { Fe } ^ { 2 + }$ in iron supplements to produce ferrous ellagic acid, which is unstable and easily oxidized to ferrous ellagic acid, so you can't drink green tea when you take iron supplements, D. Correct;

Question 13: 16. Liquid $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ is an effective oxidizer for rocket propuls...

16. Liquid $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ is an effective oxidizer for rocket propulsion systems. A small amount of $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ is prepared in the laboratory as follows, and the following analysis is incorrect ( ) $\mathrm { NH } _ { 4 } \mathrm { Cl }$ solid ${ } ^ { - - } \stackrel { \mathrm { Ca } ( \mathrm { O } ) _ { 2 } \text { 固体 } } { ( 1 ) } - \rightarrow \mathrm { NH } _ { 3 } \xrightarrow { - \frac { \text { 催化氦化 } } { ( 2 ) } } \mathrm { NO } ^ { - - } \xrightarrow { - \frac { \mathrm { O } } { ( 3 ) } } \rightarrow \mathrm { NO } _ { 2 } ( \text { 红棕色 } ) ^ { - - \frac { \text { 降温 } } { ( 4 ) } } \rightarrow \rightarrow \mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ liquid (colorless) ![](/images/questions/industrial-chemistry/image-007.jpg)

  • A. A. The valence of nitrogen changes in reactions (2) and (3)
  • B. B. Reaction (2) utilizes the reducing properties of $\mathrm { NH } _ { 3 }$
  • C. C. The color change in reaction (4) is caused by a shift in chemical equilibrium
  • D. D. Reaction (1) can be realized by the device shown in the figure

Answer: D

Solution: A. It is known that the valence of the N element in these five compounds $\mathrm { NH } _ { 4 } \mathrm { Cl } , \mathrm { NH } _ { 3 } , \mathrm { NO } , \mathrm { NO } _ { 2 } , \mathrm {~N} _ { 2 } \mathrm { O } _ { 4 }$ is -3, $- 3 , + 2 , + 4 , + 4$ respectively, so the valence of the N element in the reaction (1) (4) has not changed, and the valence of the N element in the reaction (2) (3) has changed, so A is correct; B. In reaction (2), the N element increases from -3 to +2 valence, $\mathrm { NH } _ { 3 }$ showing reduction, so B is correct; C. Since $2 \mathrm { NO } _ { 2 } \rightleftharpoons \mathrm {~N} _ { 2 } \mathrm { O } _ { 4 }$ is an exothermic reaction, the equilibrium is shifted positively by lowering the temperature so that $\mathrm { NO } _ { 2 }$ is converted to $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$, so C is correct; D. Reaction (1) is a solid heated to produce water-soluble $\mathrm { NH } _ { 3 }$, should be solid + solid heating to prepare the gas generator, the downward exhaust method to collect ammonia; the device shown in the illustration of the generator and the collection device are not appropriate, so D is not correct; The answer is D.

Question 14: 17. Sodium borohydride is a white powder, commonly used as a reducing agent in organic synthesis, an...

17. Sodium borohydride is a white powder, commonly used as a reducing agent in organic synthesis, and decomposes in moist air. Sodium metaborate ( $\mathrm { NaBO } _ { 2 }$) is soluble in water, insoluble in ethanol, easily hydrolyzed, but can be stabilized under alkaline conditions. The procedure for making $\mathrm { NaBH } _ { 4 }$ from boron concentrate (which contains a certain amount of $\mathrm { B } _ { 2 } \mathrm { O } _ { 3 }$, and impurities such as $\mathrm { Al } _ { 2 } \mathrm { O } _ { 3 } , \mathrm { SiO } _ { 2 } , \mathrm { Fe } _ { 2 } \mathrm { O } _ { 3 }$), and analyzing the purity of a sample of $\mathrm { NaBH } _ { 4 }$ (with the impurity not taking part in the reaction) is as follows. ![](/images/questions/industrial-chemistry/image-008.jpg) Step I : Take $\mathrm { mg } _ { \text {样品配成 } } 250 \mathrm {~mL} _ { \text {溶液 } }$, take ${ } ^ { 25.00 m L }$ in an iodine measuring flask, add $\mathrm { V } _ { 1 } \mathrm { ml } \mathrm { c } _ { 1 } \mathrm {~mol} \cdot \mathrm {~L}$ of $\mathrm { V } _ { 1 } \mathrm { ml } \mathrm { c } _ { 1 } \mathrm {~mol} \cdot \mathrm {~L}$ to it. $\mathrm { KIO } _ { 3 \text { 溶液,发生反应:} } \mathrm { NaBH } _ { 4 } + \mathrm { KIO } _ { 3 } \rightarrow \mathrm { NaBO } _ { 2 } + \mathrm { KI } + \mathrm { H } _ { 2 } \mathrm { O }$ (the chemical equation is not leveled). Step II: Add excess ${ } ^ { K I }$ solution to the solution from the reaction in Step I. Adjust ${ } ^ { p H }$ so that $\mathrm { KIO } ^ { 3 }$ is converted to $\mathrm { I } _ { 2 }$, cool it down, and then leave it in the dark for several minutes. Step III: Adjust ${ } ^ { \mathrm { pH } }$ of the mixture obtained in Step II to weak acidity, add a few drops of starch solution, titrate with ${ } ^ { \mathrm { C } _ { 2 } \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 } }$ standard solution to the end point, a reaction occurs: $\mathrm { I } _ { 2 } + 2 \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 } = 2 \mathrm { NaI } + \mathrm { Na } _ { 2 } \mathrm {~S} _ { 4 } \mathrm { O } _ { 6 }$, the consumption of of standard solution $\mathrm { V } _ { 2 } \mathrm {~mL}$. The following statement is correct ( )

  • A. A. The main constituents of the filtrate are $\mathrm { Fe } ( \mathrm { OH } ) _ { 3 }$
  • B. B. The main reason for adding CaO without $\mathrm { CaCl } _ { 2 }$ in the "silica-aluminum removal" step is to minimize the introduction of impurity ions and to create an alkaline environment that prevents hydrolysis of $\mathrm { NaBO } _ { 2 }$.
  • C. C. "Operation 2" is the evaporation, crystallization and washing of the filtrate, where the reagent of choice for washing is preferably ice water.
  • D. D. At the end of "operation 2" $\mathrm { NaBH } _ { 4 }$ can also be prepared by electrolysis if corrosion-resistant electrode materials are used, and the process produces ${ } ^ { 1 \mathrm {~mol} \mathrm { NaBH } } 4$ at the cathode for every ${ } ^ { 8 \mathrm {~mol} }$ electron transferred.

Answer: B

Solution: A. Iron oxide does not react with water and sodium hydroxide solution, the main component of the filtrate is $\mathrm { Fe } _ { 2 } \mathrm { O } _ { 3 } , \mathrm {~A}$ wrong; B. It is known that sodium metaborate ( $\mathrm { NaBO } _ { 2 }$) is soluble in water, insoluble in ethanol, easy to hydrolyze, but can be stabilized under alkaline conditions, then the main reason for adding CaO instead of adding $\mathrm { CaCl } _ { 2 }$ in the step of "removing silica-aluminum" is to try not to introduce impurity ions and create an alkaline environment to prevent $\mathrm { Fe } _ { 2 } \mathrm { O } _ { 3 } , \mathrm {~A}$. The main reason for adding CaO instead of $\mathrm { CaCl } _ { 2 }$ is to minimize the introduction of impurity ions and to create an alkaline environment to prevent hydrolysis of $\mathrm { NaBO } _ { 2 }$; C. Add quicklime to remove silica and aluminum, and after filtration, the filtrate containing mainly sodium metaborate is obtained, so operation 2 is to evaporate, crystallize, and filter the filtrate, and because sodium metaborate ($\mathrm { NaBO } _ { 2 }$) is soluble in water and insoluble in alcohol, it is washed with alcohol, so operation 2 is incorrect, and operation 2 is to evaporate, crystallize, and filter the filtrate; D. If the method of electrolysis to prepare $\mathrm { NaBH } _ { 4 }$, if the cathode only $\mathrm { BO } ^ { - }$ direct electron discharge into $\mathrm { BH } _ { 4 } ^ { - }$, the electrode reaction formula for $\mathrm { BO } _ { 2 } ^ { - } + 8 \mathrm { e } ^ { - } + 6 \mathrm { H } _ { 2 } \mathrm { O } = \mathrm { BH } _ { 4 } ^ { - } + 8 \mathrm { OH } ^ { - }$, the anode for $2 \mathrm { H } _ { 2 } \mathrm { O } - 4 \mathrm { e } ^ { - } = \mathrm { O } _ { 2 } + 4 \mathrm { H } ^ { + }$. FORMULA_9]], if each transfer $8 \mathrm {~mol} ^ { \text {电子,在阴 } }$ pole consumes $1 \mathrm {~mol} ^ { \mathrm { BO } ^ { - } }$ to generate $1 \mathrm {~mol} ^ { \mathrm { BH } _ { 4 } ^ { - } }$, the ions need to move in order to form a closed loop, and there is no ion-exchange membrane for the sodium ions to move, so it is not possible to generate $1 \mathrm {~mol} ^ { \mathrm { BH } _ { 4 } ^ { - } }$ at the cathode. INLINE_FORMULA_13]] at the cathode, not to mention that hydrogen ions generated by water ionization at the cathode may also discharge, D is wrong; B.

Question 15: 18. Waste cans (containing Al and small amounts of $\mathrm { Fe } , \mathrm { Mg }$ impurities) to ...

18. Waste cans (containing Al and small amounts of $\mathrm { Fe } , \mathrm { Mg }$ impurities) to prepare alum, analyze the process, the correct ( ) ![](/images/questions/industrial-chemistry/image-009.jpg)

  • A. A. To minimize the introduction of impurities, the NaOH solution should be replaced with KOH or ammonia.
  • B. B. "Sedimentation" as $\mathrm { Al } _ { 2 } \left( \mathrm { CO } _ { 3 } \right) _ { 3 }$
  • C. C. The above process can use excess $\mathrm { CO } _ { 2 }$ instead of $\mathrm { NH } _ { 4 } \mathrm { HCO } _ { 3 }$.
  • D. D. The main components of the filter residue are $\mathrm { Fe } ( \mathrm { OH } ) _ { 3 }$ and $\mathrm { Mg } ( \mathrm { OH } ) _ { 2 }$

Answer: C

Solution: A. Aluminum reacts with sodium hydroxide solution and potassium hydroxide solution, but not with ammonia, a weak base, so ammonia cannot be used instead of sodium hydroxide solution when dissolving cans, so A is wrong; B. From the analysis, the precipitate is aluminum hydroxide, so B is wrong; C. Carbon dioxide can also react with sodium metaaluminate solution to produce aluminum hydroxide precipitate, then you can use excess carbon dioxide instead of ammonium bicarbonate to get aluminum hydroxide precipitate, so C is correct; D. From the analysis, we can see that the main components of the filtrate are iron and magnesium, so D is wrong;

Question 16: 19. laboratory to investigate the preparation of potassium permanganate greening program, the experi...

19. laboratory to investigate the preparation of potassium permanganate greening program, the experimental process is shown in the figure: ![](/images/questions/industrial-chemistry/image-010.jpg) It is known that $\mathrm { K } _ { 2 } \mathrm { MnO } _ { 4 }$ under acidic, neutral and weakly basic environments, $\mathrm { MnO } _ { 4 } ^ { 2 - }$ undergoes its own redox (disproportionation) reaction to form $\mathrm { MnO } _ { 4 } ^ { - }$ and $\mathrm { MnO } _ { 2 }$. The following statement is incorrect ( )

  • A. A. The "eutectic" feeding sequence is: KOH is added first, then $\mathrm { MnO } _ { 2 }$ and $\mathrm { KClO } _ { 3 }$ are added after melting.
  • B. B. The $\mathrm { CH } _ { 3 } \mathrm { COOH }$ added during "disproportionation" provides an acidic environment while being oxidized to $\mathrm { CO } _ { 2 }$.
  • C. C. The ratio of the amount of the oxidized product to the amount of the reduced product from the "disproportionation" step is $2 : 1$
  • D. D. From the point of view of improving the utilization of raw materials, the advantage of this experimental process is that the filter residue can be recycled

Answer: B

Solution: A. $\mathrm { KClO } _ { 3 }$ is easy to decompose by heat, so it should be added at the end. The melting point of $\mathrm { MnO } _ { 2 }$ is higher than that of $\left( 535 ^ { \circ } \mathrm { C } \right)$ and the melting point is higher than that of $\mathrm { KOH } ($, so it can save energy by adding KOH first. Therefore, the feeding order of "eutectic" is: add KOH first, then add $\mathrm { MnO } _ { 2 }$ and $\mathrm { KClO } _ { 3 }$ after melting, so A is correct; B. The role of $\mathrm { CH } _ { 3 } \mathrm { COOH }$ in the process of "disproportionation" is to adjust the $\mathrm { pH } , \mathrm { K } _ { 2 } \mathrm { MnO } _ { 4 }$ of the solution to get the oxidized product $\left( \mathrm { KMnO } _ { 4 } \right)$ and the reduced product $\left( \mathrm { MnO } _ { 2 } \right)$. _10]], the added $\mathrm { CH } _ { 3 } \mathrm { COOH }$ will not be oxidized by the generated $\mathrm { KMnO } _ { 4 }$, so item B is wrong; C. The $\mathrm { K } _ { 2 } \mathrm { MnO } _ { 4 }$ reaction in the "disproportionation" step yields the oxidized product $\left( \mathrm { KMnO } _ { 4 } \right)$ and the reduced product $\left( \mathrm { MnO } _ { 2 } \right)$, and a portion of the $\mathrm { K } _ { 2 } \mathrm { MnO } _ { 4 }$ is oxidized by the $\mathrm { K } _ { 2 } \mathrm { MnO } _ { 4 }$. 16]], one part $\mathrm { K } _ { 2 } \mathrm { MnO } _ { 4 }$ in which the valence of the element Mn is increased by 1 to produce $\mathrm { KMnO } _ { 4 }$, and the other part $\mathrm { K } _ { 2 } \mathrm { MnO } _ { 4 }$ in which the valence of the element Mn is decreased by 2 to produce $\mathrm { MnO } _ { 2 }$, and according to the conservation of gain and loss of electrons, the ratio of the amount of substance of oxidized product to that of the reduced product is [$\mathrm { MnO } _ { 2 }$, which is $\left( \mathrm { MnO } _ { 2 } \right)$. INLINE_FORMULA_20]], so C is correct; D. According to the analysis, the filtrate obtained in this step is $\mathrm { MnO } _ { 2 }$, which can be recycled, so D is correct; The answer is B.

Question 17: 20. Liquid $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ is an effective oxidizer for rocket propuls...

20. Liquid $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ is an effective oxidizer for rocket propulsion systems. A small amount of $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ is prepared in the laboratory as follows. The following analysis is incorrect ( ) $\underset { \text { 固体 } } { \mathrm { NH } _ { 4 } \mathrm { Cl } } \xrightarrow [ \text {(1)} ] { \mathrm { Ca } ( \mathrm { OH } ) _ { 2 } \text { 固体 } } \mathrm { NH } _ { 3 } \xrightarrow [ \text {(2)} ] { \text { 催化氧化 } } \mathrm { NO } \xrightarrow [ \text {(3)} ] { \mathrm { O } _ { 2 } } \underset { \text {(红棕色)} } { \mathrm { NO } _ { 2 } } \xrightarrow [ \text {(4)} ] { \text { 降温 } } \underset { \text {(无色)} } { \mathrm { N } _ { 2 } \mathrm { O } _ { 4 } \text { 液体 } }$ ![](/images/questions/industrial-chemistry/image-011.jpg)

  • A. A. The valence of nitrogen changes in reactions (2) and (3)
  • B. B. Reaction (2) utilizes the reducing properties of $\mathrm { NH } _ { 3 }$
  • C. C. The color change in reaction (4) is caused by a shift in chemical equilibrium
  • D. D. Reaction (1) can be realized by the device shown in the figure

Answer: D

Solution: A. It is known that the valence of the N element in these five compounds $\mathrm { NH } _ { 4 } \mathrm { Cl } , \mathrm { NH } _ { 3 } , \mathrm { NO } , \mathrm { NO } _ { 2 } , \mathrm {~N} _ { 2 } \mathrm { O } _ { 4 }$ is -3, -3, +2, +4, +4, so the valence of the N element in the reaction (1) (4) has not changed, and the valence of the N element in the reaction (2) (3) has changed, so A is correct; B. In reaction (2), the N element increases from -3 to +2 valence, and $\mathrm { NH } _ { 3 }$ shows reduction, so B is correct; C. Since $2 \mathrm { NO } _ { 2 } \rightleftharpoons \mathrm {~N} _ { 2 } \mathrm { O } _ { 4 }$ is an exothermic reaction, the equilibrium is shifted positively by lowering the temperature so that $\mathrm { NO } _ { 2 }$ is converted to $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$, so C is correct; D. Reaction (1) is a solid heated to generate water-soluble $\mathrm { NH } _ { 3 }$, should be solid + solid heating to prepare the gas generator, downward exhaust method to collect ammonia; the device in the illustration of the generator and the collection device are not appropriate, so the D is incorrect; D. The answer is selected.

Question 18: 21. One of the methods of industrial mercury production is cinnabar (containing mainly HgS) roasting...

21. One of the methods of industrial mercury production is cinnabar (containing mainly HgS) roasting, and the main reaction that occurs when cinnabar is roasted is The main reaction that occurs when cinnabar is roasted is $\mathrm { HgS } + \mathrm { O } _ { 2 } \xlongequal { \text { 焙烧 } } \mathrm { Hg } + \mathrm { SO } _ { 2 }$, which produces gaseous mercury, which is condensed to obtain liquid mercury. A process for preparing pure mercury from cinnabar is shown in the figure. The following analysis is correct ( ) ![](/images/questions/industrial-chemistry/image-012.jpg)

  • A. A. The main purpose of adding CaO during roasting is to lower the melting point of HgS
  • B. B. Mercury-containing flue gas requires mercury removal before it can be released to the atmosphere
  • C. C. Physical changes occur when washing with nitric acid in $5 \%$.
  • D. D. Distillation in the laboratory requires apparatus such as distillation flasks, condenser tubes and dispensing funnels

Answer: B

Solution: A. The main purpose of adding CaO during roasting is to solidify sulfur and reduce sulfur dioxide emissions and air pollution, so A is wrong; B. Industrial waste gas needs to be treated and discharged, so B is correct; C. washing with $5 \%$ of nitric acid is to dissolve lead, copper and other impurities, a chemical change, so C is wrong; D. distillation is a mutually soluble liquids using the boiling point of different properties of the separation of the operation, the need to use the apparatus are: distillation flask, condenser tube, etc., do not need to be divided into funnels, so D error; The answer is B. [Eyes on] add quicklime or calcium carbonate sulfur fixation, reduce sulfur dioxide gas emissions; industrial waste gas, waste water, waste residue need to be treated and discharged.

Question 19: 22. The figure below shows the process of laboratory purification of potassium chloride containing a...

22. The figure below shows the process of laboratory purification of potassium chloride containing a small amount of barium chloride impurity. The following analysis is incorrect () ![](/images/questions/industrial-chemistry/image-013.jpg)

  • A. A. Operation I is filtration to remove the solids.
  • B. B. Reagent X is $\mathrm { K } _ { 2 } \mathrm { CO } _ { 3 }$ solution and operation II is filtration.
  • C. C. Add reagent X in excess
  • D. D. The reagent $Y$ is dilute hydrochloric acid.

Answer: A

Solution: A, the purification of potassium chloride containing a small amount of barium chloride impurities, because barium chloride and potassium chloride are soluble in water, so operation 1 is to use the appropriate amount of water will be dissolved, option A is incorrect; B, the reagent X is $\mathrm { K } _ { 2 } \mathrm { CO } _ { 3 }$ solution, potassium carbonate and barium chloride reaction to produce barium carbonate precipitate and potassium chloride, through the operation of the II filtration will be filtered barium carbonate precipitate filtrate, filtrate for potassium chloride, option B Correct; C. Add reagent X in excess to precipitate barium ions completely, option C is correct; D. Reagent Y is dilute hydrochloric acid, which reacts with barium carbonate to form barium chloride, restoring the purified substance, option D is correct. The answer is A.

Question 20: 23. Sodium borohydride is a white powder, commonly used as a reducing agent in organic synthesis, an...

23. Sodium borohydride is a white powder, commonly used as a reducing agent in organic synthesis, and decomposes in moist air. Sodium metaborate ( $\mathrm { NaBO } _ { 2 }$) is soluble in water, insoluble in ethanol, easily hydrolyzed, but can be stabilized under alkaline conditions. The procedure for making $\mathrm { NaBH } _ { 4 }$ from boron concentrate (which contains a certain amount of $\mathrm { B } _ { 2 } \mathrm { O } _ { 3 }$, and impurities such as $\mathrm { Al } _ { 2 } \mathrm { O } _ { 3 } , \mathrm { SiO } _ { 2 } , \mathrm { Fe } _ { 2 } \mathrm { O } _ { 3 }$), and analyzing the purity of a sample of $\mathrm { NaBH } _ { 4 }$ (with the impurity not taking part in the reaction) is as follows. ![](/images/questions/industrial-chemistry/image-014.jpg) Step I : Take $\mathrm { mg } _ { \text {样品配成 } } 250 \mathrm {~mL} _ { \text {溶液 } }$, take $25.00 \mathrm {~mL} _ { \text {置于碘量瓶中,add } } \mathrm { V } _ { 1 } \mathrm { ml } \mathrm { c } _ { 1 } \mathrm {~mol} \cdot \mathrm {~L}$'s $\mathrm { KIO } _ { 3 \text { 溶液,发生反应:} } \mathrm { NaBH } _ { 4 } + \mathrm { KIO } _ { 3 } \rightarrow \mathrm { NaBO } _ { 2 } + \mathrm { KI } + \mathrm { H } _ { 2 } \mathrm { O }$ (the chemical equation is not leveled). Step II: Add excess $\mathrm { KI } _ { \text {溶液,调节 } } \mathrm { pH }$ to the solution from the reaction in Step I to convert $\mathrm { KIO } ^ { 3 }$ to $\mathrm { I } _ { 2 }$, cool, and leave in the dark for a few minutes. Step III: Adjust ${ } ^ { \mathrm { pH } }$ of the mixture obtained in Step II to weak acidity, add a few drops of starch solution, titrate with ${ } ^ { \mathrm { C } _ { 2 } } \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 }$ standard solution to the end point, a reaction occurs: $\mathrm { I } _ { 2 } + 2 \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 } = 2 \mathrm { NaI } + \mathrm { Na } _ { 2 } \mathrm {~S} _ { 4 } \mathrm { O } _ { 6 }$, consume $\mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 }$ standard solution $\mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 }$ and $\mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 }$. _14]] standard solution $\mathrm { V } _ { 2 } \mathrm {~mL}$ The purity of the sample is correctly calculated as

  • A. A. $\frac { 15 \left( c _ { 1 } V _ { 1 } - \frac { 1 } { 6 } c _ { 2 } V _ { 2 } \right) } { 2 m } \%$
  • B. B. $\frac { 57 \left( c _ { 1 } V _ { 1 } - \frac { 1 } { 2 } c _ { 2 } V _ { 2 } \right) } { 2 m } \%$
  • C. C. $\frac { 57 \left( c _ { 1 } V _ { 1 } - \frac { 1 } { 6 } c _ { 2 } V _ { 2 } \right) } { 2 m } \%$
  • D. D. $\frac { 15 \left( c _ { 1 } V _ { 1 } - \frac { 1 } { 2 } c _ { 2 } V _ { 2 } \right) } { 2 m } \%$

Answer: C

Solution: According to the question, the amount of substance of the standard solution consumed ${ } ^ { \mathrm { c } _ { 2 } } \mathrm {~V} _ { 2 } \times 10 ^ { - 3 } \mathrm {~mol}$, by the reaction of $\mathrm { I } _ { 2 } + 2 \mathrm { Na } _ { 2 } \mathrm {~S} _ { 2 } \mathrm { O } _ { 3 } = 2 \mathrm { NaI } + \mathrm { Na } _ { 2 } \mathrm {~S} _ { 4 } \mathrm { O } _ { 6 }$, according to $\mathrm { I } _ { 2 } \sim 2 \mathrm { NaS } _ { 2 } \mathrm { O } _ { 3 }$, the amount of substance of iodine is $\frac { \mathrm { c } _ { 2 } \mathrm {~V} _ { 2 } \times 10 ^ { - 3 } } { 2 } \mathrm {~mol}$, according to [[INLINE_FORMULA According to $\mathrm { KIO } _ { 3 } + 5 \mathrm { KI } + 3 \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } = 3 \mathrm {~K} _ { 2 } \mathrm { SO } _ { 4 } + 3 \mathrm { I } _ { 2 } + 3 \mathrm { H } _ { 2 } \mathrm { O }$, the amount of potassium iodate reacting with potassium iodide is $\frac { \mathrm { C } _ { 2 } \mathrm {~V} _ { 2 } \times 10 ^ { - 3 } } { 6 } \mathrm {~mol}$, and the amount of potassium iodate reacting with $\mathrm { NaBH } _ { 4 }$ is $\left( c _ { 1 } V _ { 1 } - \frac { 1 } { 6 } c _ { 2 } V _ { 2 } \right) \times 10 ^ { - 3 } \mathrm {~mol} , 3 \mathrm { NaBH } _ { 4 } \sim 4 \mathrm { KIO } _ { 3 }$, and the amount of iodine reacting with potassium iodide is $\mathrm { NaBH } _ { 4 }$. According to $\left( c _ { 1 } V _ { 1 } - \frac { 1 } { 6 } c _ { 2 } V _ { 2 } \right) \times 10 ^ { - 3 } \mathrm {~mol} , 3 \mathrm { NaBH } _ { 4 } \sim 4 \mathrm { KIO } _ { 3 }$, the amount of potassium iodate that reacts with $\mathrm { NaBH } _ { 4 }$ is $\left( c _ { 1 } V _ { 1 } - \frac { 1 } { 6 } c _ { 2 } V _ { 2 } \right) \times 10 ^ { - 3 } \mathrm {~mol} , 3 \mathrm { NaBH } _ { 4 } \sim 4 \mathrm { KIO } _ { 3 }$, then the amount of $\mathrm { NaBH } _ { 4 }$ is $\frac { 3 } { 4 } \left( c _ { 1 } V _ { 1 } - \frac { 1 } { 6 } c _ { 2 } V _ { 2 } \right) \times 10 ^ { - 3 } \mathrm {~mol}$, and the mass fraction of $\mathrm { NaBH } _ { 4 }$ in the sample is $$ \frac { \frac { 3 } { 4 } \left( c _ { 1 } V _ { 1 } - \frac { 1 } { 6 } c _ { 2 } V _ { 2 } \right) \times 10 ^ { - 3 } \mathrm {~mol} \times 38 \mathrm {~g} / \mathrm { mol } \times \frac { 250 \mathrm {~mL} } { 25.00 \mathrm {~mL} } } { \mathrm { mg } } \times 100 \% = \frac { 57 \left( c _ { 1 } V _ { 1 } - \frac { 1 } { 6 } c _ { 2 } V _ { 2 } \right) } { 2 m } \% $$ Option C.

Question 21: 24. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic aci...

24. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid solution is colorless, ferric ellagic acid solution is blue-black. According to the analysis of the experimental process shown in the figure, the following description is wrong $\underset { \text { 叶水 } } { \text { 绿茶 } } \underset { \text {(1)add } } { \mathrm { FeSO } _ { 4 } }$ blue-black $\underset { \text {(2)add } } { \text { 维生素 } } \underset { \text { 原色 } } { \text { 恢复为 } } \underset { \text {(3)drop in } } { \mathrm { KSCN } \text { 溶液 } }$ no color change $\xrightarrow [ \text {(4)drop in } ] { \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } \text { 溶液 } } \underset { \text { 红色 } } { \text { 变为 } }$

  • A. A. Reaction (1) in which green tea leaf water reacts directly with $\mathrm { FeSO } _ { 4 }$ to form iron tannate which is blue-black in color
  • B. B. Vitamin C in reaction (2) exhibits reducing properties
  • C. C. In reaction (4) $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ can be replaced with freshly made chlorine water.
  • D. D. Green tea leaf water should not be taken at the same time as iron supplements

Answer: A

Solution: A. Green tea contains a large amount of ellagic acid and $\mathrm { FeSO } _ { 4 }$ reacts with $\mathrm { FeSO } _ { 4 }$ to form ferrous ellagic acid, the solution is colorless, ferrous ellagic acid is oxidized to form ferric ellagic acid and is blue-black, A is wrong; B. Add vitamin C to the blue-black solution, and the solution returns to its original color from blue-black, indicating that vitamin C can reduce iron ellagic acid to ferrous ellagic acid, which is reducing, B is correct; C. Both $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ and freshly made chlorine water have oxidizing property, which can oxidize $\mathrm { Fe } ^ { 2 + }$ to $\mathrm { Fe } ^ { 3 + }$, so $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ can be replaced by freshly made chlorine water in reaction (4), C is correct; D. Because the ellagic acid in green tea water and iron supplements in the $\mathrm { Fe } ^ { 2 + }$ reaction to generate ferrous ellagic acid, ferrous ellagic acid is unstable, easy to be oxidized to ferric ellagic acid, so you can not drink green tea when taking iron supplements, D is correct; Therefore, the answer is A.

Question 22: 25. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic aci...

25. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid solution is colorless, ferric ellagic acid solution is blue-black, generalized as shown in the figure of the experimental process analysis, the following description is wrong $\underset { \text { 叶水 } } { \text { 绿 } } \underset { \text {(1)add } } { \mathrm { FeSO } _ { 4 } } \underset { \text { 蓝黑色 } } { \text { Vitamin C } } \underset { \text {(2)add } } { \text { 恢复为 } } \underset { \text {(3)drop in } } { \mathrm { KSCN } \text { 溶液 } } \underset { \text {(4)drop in } } { \text { 不变色 } } \xrightarrow [ \text { 红色 } ] { \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } \text { 溶液 } }$

  • A. A. The ferrous ellagic acid produced by the reaction of green tea leaf water with $\mathrm { FeSO } _ { 4 }$ is easily oxidized to ferric ellagic acid, which is blue-black in color.
  • B. B. Vitamin C reduces ferric ellagic acid to ferrous ellagic acid
  • C. C. (2 reactions occur in (4)
  • D. D. You should not drink green tea when taking iron supplements

Answer: C

Solution: A. Ferrous ellagic acid solution is colorless, ferric ellagic acid solution is blue-black, add FeSO 4 to green tea water, blue-black, indicating that ferrous ellagic acid is easily oxidized, A is correct; B. To the blue-black solution, add vitamin C, the solution from blue-black color back to the original color, indicating that vitamin C can reduce iron ellagic acid into ferrous ellagic acid, the performance of reducing, B is correct; C. The reactions in (4) include vitamin C, oxidation of $\mathrm { Fe } ^ { 2 + }$ by $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$, and complexation between $\mathrm { Fe } ^ { 3 + }$ and $\mathrm { SCN } ^ { - }$. ## C is incorrect; D. Because the ellagic acid in green tea leaf water reacts with $\mathrm { Fe } ^ { 2 + }$ in iron supplements to produce ferrous ellagic acid, which is unstable and easily oxidized to ferrous ellagic acid, so you can't drink green tea when you take iron supplements, D. Correct; Therefore, the answer is C.

Question 23: 28. Cadmium ferrite (mainly composed of $\mathrm { CdO } _ { 2 } , \mathrm { Fe } _ { 2 } \mathrm { ...

28. Cadmium ferrite (mainly composed of $\mathrm { CdO } _ { 2 } , \mathrm { Fe } _ { 2 } \mathrm { O } _ { 3 } , \mathrm { Fe }$ and $\mathrm { SiO } _ { 2 }$) is used as a raw material for the preparation of analytical reagents $\mathrm { Cd } \left( \mathrm { BrO } _ { 3 } \right) _ { 2 }$. The process of $\mathrm { Cd } \left( \mathrm { BrO } _ { 3 } \right) _ { 2 }$ is shown in the figure $\left[ \mathrm { Cd } \left( \mathrm { SO } _ { 4 } \right) _ { 2 } \right.$ is soluble in water $]$: ![](/images/questions/industrial-chemistry/image-015.jpg) The following statement is incorrect

  • A. A. The acid leaching process involves both redox and non-redox reactions.
  • B. B. $\mathrm { CH } _ { 3 } \mathrm { OH }$ is a reducing agent in the cadmium reduction step, and $\mathrm { Cd } ^ { 4 + }$ in solution may oxidize it to $\mathrm { CO } _ { 2 }$
  • C. C. If bubbles are generated in the oxidation step, it means that ${ } ^ { \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } }$ can also oxidize $\mathrm { CH } _ { 3 } \mathrm { OH }$ under acidic conditions.
  • D. D. A small amount of $\mathrm { Cd } ^ { 2 + }$ remains in the filtrate after cadmium precipitation, which needs to be further recycled before discharge to prevent environmental pollution.

Answer: C

Solution: A. Acid leaching process $\mathrm { CdO } _ { 2 } , \mathrm { Fe } _ { 2 } \mathrm { O } _ { 3 }$ and dilute sulfuric acid complex decomposition reaction generates salt and water, belongs to the non-redox reaction, Fe and sulfuric acid reaction generates ferrous sulfate and hydrogen belongs to the redox reaction, option A is correct; B. $\mathrm { CH } _ { 3 } \mathrm { OH }$ is the reducing agent in the step of reducing cadmium, and $\mathrm { Cd } ^ { 4 + }$ in the solution may oxidize it to $\mathrm { CO } _ { 2 } , \mathrm { Cd } ^ { 4 + }$ which is converted to $\mathrm { Cd } ^ { 2 + }$, and option B is correct; C. The bubbles may be produced by the decomposition of hydrogen peroxide catalyzed by $\mathrm { Cd } ^ { 2 + }$, etc., option C is incorrect; D. Cadmium is a heavy metal, and a small amount of $\mathrm { Cd } ^ { 2 + }$ will remain in the filtrate after cadmium precipitation, which needs to be further recycled before discharge to prevent pollution of the environment, option D is correct; Answer choice C.

Question 24: 29. The electronics industry uses $\mathrm { FeCl } _ { 3 }$ solutions to etch copper foil from insu...

29. The electronics industry uses $\mathrm { FeCl } _ { 3 }$ solutions to etch copper foil from insulating boards to manufacture printed circuit boards. The process for recovering copper and ${ } ^ { \mathrm { FeCl } _ { 3 } }$ solids from the acid etching waste solution is as follows: ![](/images/questions/industrial-chemistry/image-016.jpg) Known: $\mathrm { SOCl } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { O } \square$ convex $\mathrm { SO } _ { 2 } \uparrow + 2 \mathrm { HCl } \uparrow$ The following analysis is incorrect

  • A. A. Gases are produced in processes I, II, III, and V.
  • B. B. The reactions that take place in the above process are replacement reactions, chemical reactions
  • C. C. $V$ in $\mathrm { SOCl } _ { 2 }$ instead of direct heating and dehydration, mainly to avoid $\mathrm { FeCl } _ { 3 }$ hydrolysis.
  • D. D. Add hydrochloric acid to II until no more bubbles are produced, then add ${ } ^ { \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } }$ solution to the solution obtained by filtration to increase the $\mathrm { FeCl } _ { 3 }$ yield.

Answer: D

Solution: A. In process | iron reacts with acid to produce hydrogen, in II the iron in the slag reacts with hydrochloric acid to produce hydrogen, in III hydrogen peroxide oxidizes ferrous chloride to ferric chloride, $\mathrm { Fe } ^ { 3 + }$ is a catalyst for the decomposition of $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$, and there will be the generation of oxygen, and in V $\mathrm { SOCl } _ { 2 }$ interacts with the crystalline water in ferric chloride crystal to produce gas $\mathrm { SOCl } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { O } = \mathrm { SO } _ { 2 } \uparrow + 2 \mathrm { HCl } \uparrow$, so A is correct; therefore, A is correct. In V $\mathrm { SOCl } _ { 2 }$, a gas $\mathrm { SOCl } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { O } = \mathrm { SO } _ { 2 } \uparrow + 2 \mathrm { HCl } \uparrow$ is produced by interaction with the water of crystallization in ferric chloride crystals, so A is correct; B. Reactions occurring in the above process include replacement reactions such as the reaction between iron and hydrochloric acid, and chemical reactions such as the reaction between ferric chloride and iron, so B is correct; C. In V, $\mathrm { SOCl } _ { 2 }$ is used instead of direct heating and dehydration, because the reaction $\mathrm { SOCl } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { O } = \mathrm { SO } _ { 2 } \uparrow + 2 \mathrm { HCl } \uparrow$ occurs to inhibit the hydrolysis of $\mathrm { FeCl } _ { 3 }$, so C is correct; D. In II, hydrochloric acid is added dropwise until no gas escapes, which means that the hydrochloric acid reacts completely, and the solution obtained by filtration is added to $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ solution, which is needed to oxidize the ferrous chloride to ferric chloride under acidic conditions, which improves the yield of $\mathrm { FeCl } _ { 3 }$, so D is wrong;

Question 25: 30. $\mathrm { MnO } _ { 2 }$ is an important inorganic functional material. The process of producin...

30. $\mathrm { MnO } _ { 2 }$ is an important inorganic functional material. The process of producing $\mathrm { MnO } _ { 2 }$ from crude $\mathrm { MnO } _ { 2 }$ (which contains the impurities MnO and $\mathrm { MnCO } _ { 3 }$) is shown in the figure. ![](/images/questions/industrial-chemistry/image-017.jpg) The following analysis is incorrect

  • A. A. The name of operation X and operation Y are both filtered.
  • B. B. $\mathrm { CO } _ { 2 }$ is an acidic oxide, $\mathrm { MnO } _ { 2 }$ is a basic oxide.
  • C. C. The ratio of the amount of substance of the oxidizing agent to that of the reducing agent in the process "oxidation" is $2 : 5$
  • D. D. $\mathrm { Cl } _ { 2 }$ reacts with NaOH solution, transferring electrons per $42.6 \mathrm {~g} \mathrm { NaClO } _ { 3 }$ to $2 N _ { A }$

Answer: B

Solution:

Question 26: 31. In the industrial treatment of zinc smelting residues from sphalerite (containing mainly $\mathr...

31. In the industrial treatment of zinc smelting residues from sphalerite (containing mainly $\mathrm { Zn } , \mathrm {~Pb} , \mathrm { Fe } , \mathrm { Ga }$ and other elements), the following process is designed for the extraction of lead, zinc and gallium. Known: $\mathrm { Ga } ( \mathrm { OH } ) _ { 3 } + \mathrm { OH } ^ { - } = \left[ \mathrm { Ga } ( \mathrm { OH } ) _ { 4 } \right] ^ { - }$ ![](/images/questions/industrial-chemistry/image-018.jpg) The process is incorrectly analyzed by

  • A. A. Slag 1 is composed of Pb monomers.
  • B. B. Reagent a is ZnO
  • C. C. The main constituents of sludge 3 are $\mathrm { Fe } ( \mathrm { OH } ) _ { 3 }$
  • D. D. The cathodic reaction of filtrate 3 undergoing electrolysis is $\left[ \mathrm { Ga } ( \mathrm { OH } ) _ { 4 } \right] ^ { - } + 3 \mathrm { e } ^ { - } = \mathrm { Ga } ^ { - } + 4 \mathrm { OH } ^ { - }$

Answer: A

Solution: A. Slag 1 is lead sulfate, so A is wrong; B. Reagent a is ZnO, which can be used to adjust pH and does not introduce new impurities, so B is correct; C. Iron hydroxide does not react with sodium hydroxide, so the main component of slag 3 is $\mathrm { Fe } ( \mathrm { OH } ) _ { 3 }$, so C is correct; D. Electrolysis $\left[ \mathrm { Ga } ( \mathrm { OH } ) _ { 4 } \right] ^ { - }$ of gallium, cathode reduction reaction, cathode electrode reaction formula: $\left[ \mathrm { Ga } ( \mathrm { OH } ) _ { 4 } \right] ^ { - } + 3 \mathrm { e } ^ { - } = \mathrm { Ga } ^ { - } + 4 \mathrm { OH } ^ { - }$, so D is correct.

Question 27: 32. The method of treating malaria contained in the Chinese medical text "After Elbow Prepared Emerg...

32. The method of treating malaria contained in the Chinese medical text "After Elbow Prepared Emergency Formula", which is "one grip of artemisia, impregnate it with two liters of water, extract the juice by twisting it, and take it as much as you like", is a great inspiration for the research and development of artemisinin, a potent drug for malaria by Tu Youyou's team. Tu Youyou discovered that the ancient method was ineffective in comparative experiments, and she designed a process for extracting artemisinin that could be simplified as follows: ![](/images/questions/industrial-chemistry/image-019.jpg) Analysis, the following judgment or narrative is certainly wrong ()

  • A. A. Extracting active ingredients from natural plants to treat diseases is an important approach to drug development
  • B. B. Crushing Artemisia annua can improve the extraction rate of active ingredients
  • C. C. Artemisinin is soluble in water and ether.
  • D. D. operation 1 is filtration, operation 2 is distillation

Answer: C

Solution: A, from natural plants to extract active ingredients to treat diseases is an important way of drug research and development, such as Chinese herbal medicine, so A is correct; B, will be crushed artemisia can increase the contact area with the extractant, improve the extraction rate of active ingredients, so B is correct; C, artemisinin is easily soluble in ether, so use ether to extract, but artemisinin is not easy to dissolve in water because of the question, "Artemisia annua a grip. A grip of artemisia, two liters of water stains, twisted juice", but in the experiment found that the ancient method of low efficacy, and its structure in the hydrophilic group is very little, that artemisinin is not easy to dissolve in water, so C is wrong; D, according to the figure in the process, the operation 1 is filtered, the operation 2 is distillation, it is correct, D; Therefore, choose C.

Question 28: 33. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic aci...

33. It is known that: (1) green tea contains a large amount of ellagic acid; (2) ferrous ellagic acid solution is colorless, ferric ellagic acid solution is blue-black. According to the analysis of the experimental process shown in the figure, the following description is wrong ( ) ![](/images/questions/industrial-chemistry/image-020.jpg)

  • A. A. The ferrous tannate produced by the reaction between green tea leaf water and ferrous sulfate is easily oxidized to ferric tannate and becomes blue-black in color.
  • B. B. Vitamin C in reaction (2) exhibits oxidizing properties
  • C. C. In reaction (4) $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ can be replaced with freshly made chlorine water.
  • D. D. Green tea leaf water should not be taken at the same time as iron supplements

Answer: B

Solution: A. Green tea contains a large amount of ellagic acid and $\mathrm { FeSO } _ { 4 }$ reaction to produce ferrous ellagic acid, the solution is colorless, ferrous ellagic acid is oxidized to produce iron ellagic acid and blue-black, A is correct; B. To the blue-black solution add vitamin C, the solution from blue-black color back to the original color, indicating that vitamin C can be iron ellagic acid also B. When vitamin C is added to a blue-black solution, the solution returns to its original color from blue-black; C. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ and freshly made chlorine water are both oxidizing, and can oxidize $\mathrm { Fe } ^ { 2 + }$ to $\mathrm { Fe } ^ { 3 + }$, so $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ in reaction (4) can be replaced by freshly made chlorine water, C is correct; D. Because the ellagic acid in green tea water and iron supplements in the $\mathrm { Fe } ^ { 2 + }$ reaction to generate ferrous ellagic acid, ferrous ellagic acid is unstable, easy to be oxidized to ferric ellagic acid, so you can not drink green tea when taking iron supplements, D is correct;

Question 29: 34. Potassium pertechnate $\left( \mathrm { K } _ { 2 } \mathrm { FeO } _ { 4 } \right)$ is an envir...

34. Potassium pertechnate $\left( \mathrm { K } _ { 2 } \mathrm { FeO } _ { 4 } \right)$ is an environmentally friendly, highly efficient, multi-functional drinking water treatment agent. It is an environmentally friendly, highly efficient and multi-functional drinking water treatment agent, which is made from waste iron filings (mainly into A process of preparing $\mathrm { K } _ { 2 } \mathrm { FeO } _ { 4 }$ from waste iron filings (the main component is Fe, containing a small amount of $\mathrm { Fe } _ { 2 } \mathrm { O } _ { 3 } , \mathrm { FeO }$, etc.) is shown in the figure. The following statement is correct ( ) ![](/images/questions/industrial-chemistry/image-021.jpg) $$ \mathrm { NaClO } > \mathrm { Na } _ { 2 } \mathrm { FeO } _ { 4 } $$

  • A. A. In the process of "acid leaching and oxidizing", the higher the temperature, the more favorable the reaction is
  • B. B. Oxidizing properties under alkaline conditions;
  • C. C. Solubility of the substance under the condition: $\mathrm { Na } _ { 2 } \mathrm { FeO } _ { 4 } < \mathrm { K } _ { 2 } \mathrm { FeO } _ { 4 }$
  • D. D. The chemical reactions involved in this process are all redox reactions

Answer: B

Solution: A. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ will decompose to form water and oxygen at higher temperatures, and higher temperatures are not suitable in the process of "acid leaching and oxidation", so A is wrong; B. Sodium hypochlorite oxidizes iron ions into $\mathrm { Na } _ { 2 } \mathrm { FeO } _ { 4 }$ under alkaline conditions, the chlorine elemental valence decreases, the iron elemental valence increases, the oxidizing agent is NaClO, and the oxidation product is $\mathrm { Na } _ { 2 } \mathrm { FeO } _ { 4 }$, the oxidizing property is $\mathrm { NaClO } > \mathrm { Na } _ { 2 } \mathrm { FeO } _ { 4 }$, so B is correct; C. In the crystallization process $\mathrm { Na } _ { 2 } \mathrm { FeO } _ { 4 }$, $\mathrm { Na } _ { 2 } \mathrm { FeO } _ { 4 }$ is added to concentrated KOH solution, and $\mathrm { K } _ { 2 } \mathrm { FeO } _ { 4 }$ crystals are precipitated, which indicates the solubility: $\mathrm { Na } _ { 2 } \mathrm { FeO } _ { 4 } > \mathrm { K } _ { 2 } \mathrm { FeO } _ { 4 }$, so C is wrong; D. In the process of crystallization, $\mathrm { Na } _ { 2 } \mathrm { FeO } _ { 4 }$ is added to concentrated KOH solution to produce $\mathrm { K } _ { 2 } \mathrm { FeO } _ { 4 }$ crystals and NaOH, no change in elemental valence, not a redox reaction, so D is wrong; Therefore, the answer is B.

Question 30: 35. Cuprous chloride $( \mathrm { CuCl } )$ is widely used in chemical, printing and dyeing, electro...

35. Cuprous chloride $( \mathrm { CuCl } )$ is widely used in chemical, printing and dyeing, electroplating and other industries. The process designed for the preparation of CuCl in industry using copper sulfide concentrate as raw material is shown in the figure: ![](/images/questions/industrial-chemistry/image-022.jpg) Known: CuCl is insoluble in ethanol, easy to hydrolyze when wet. The following statement is correct ( )

  • A. A. Step (1) roasting produces toxic gases that can make purple litmus solution first turn red and then fade, reflecting bleaching properties
  • B. B. If nitric acid is used instead of sulfuric acid in step (2), it will affect the yield of CuCl
  • C. C. Step (3) occurs as a complex decomposition reaction
  • D. D. Step (4) uses an ethanol wash and can also be washed with distilled water

Answer: B

Solution: A. Sulfur dioxide generated by roasting is a toxic gas. Sulfur dioxide reacts with water to form ${ } ^ { \mathrm { H } _ { 2 } \mathrm { SO } _ { 3 } }$, which is acidic and can turn purple All people's solution red, but not make it fade; B. If nitric acid is used instead of sulfuric acid in step (2), copper nitrate will be generated, and the $\mathrm { HNO } _ { 3 }$ contained in the resulting solution has strong oxidizing properties and can oxidize cuprous ions to copper ions, which is detrimental to the generation of cuprous chloride and affects the yield of CuCl, B. This is the correct solution; C. $\mathrm { Na } _ { 2 } \mathrm { SO } _ { 3 }$ has the reducing property, in step (3), sodium sulfite will reduce the copper ion to cuprous ion and then generate cuprous chloride precipitate with chlorine ions, which is a redox reaction, not a complex decomposition reaction, C is wrong; D. CuCl is insoluble in ethanol, easy to hydrolyze when wet, then can not use distilled water to wash, D error;

Question 31: 36. A process for the preparation of iron red using magnetite ${ } ^ { \left( \mathrm { Fe } _ { 3 }...

36. A process for the preparation of iron red using magnetite ${ } ^ { \left( \mathrm { Fe } _ { 3 } \mathrm { O } _ { 4 } \right.}$ is as follows: ![](/images/questions/industrial-chemistry/image-023.jpg) The following statement is correct ( )

  • A. A. A small amount of acidic $\mathrm { KMnO } _ { 4 }$ solution is added dropwise to filtrate A, and the purple color fades.
  • B. B. The presence of $\mathrm { Fe } ^ { 2 + }$ in filtrate A can be tested with KSCN solution and chlorine water.
  • C. C. $\mathrm { FeCO } _ { 3 }$ The reaction equation for calcination in air is $\mathrm { FeCO } _ { 3 } \xlongequal { \text { 煅烧 } } \mathrm { FeO } + \mathrm { CO } _ { 2 } \uparrow$
  • D. D. The mass of Fe in the final iron red is less than the mass of Fe in the original magnetite.

Answer: A

Solution: A. It can be seen from the analysis, filtrate A for iron sulfate, ferrous sulfate mixed solution, add a small amount of acidic potassium permanganate solution, ferrous sulfate solution can react with acidic potassium permanganate solution to make the solution discoloration, A is correct; B. From the analysis, we can see that the filtrate A is a mixed solution of ferrous sulfate and ferrous sulfate, so the ferrous ions in the solution can not be tested with potassium thiocyanate solution and chlorine water, B is wrong; C. From the analysis, ferrous carbonate calcined in air to generate iron oxide, the chemical equation of the reaction is $4 \mathrm { FeCO } _ { 3 } + \mathrm { O } _ { 2 } \xlongequal { \text { 煅烧 } } 2 \mathrm { Fe } _ { 2 } \mathrm { O } _ { 3 } + 4 \mathrm { CO } _ { 2 }$, C error; D. From the analysis, we can see that in the experimental process added iron, the solution of iron sulfate into ferrous sulfate, by the conservation of iron atoms can be known, the final iron oxide in the mass of iron is greater than the original magnetite in the mass of iron elements, D error; Therefore, choose A.

Question 32: 37. China's chemist Hou Debang proposed the famous Hou's alkali method, which has made outstanding c...

37. China's chemist Hou Debang proposed the famous Hou's alkali method, which has made outstanding contributions to the world's alkali industry. An experimental group simulated the process of Hou's alkali production method as shown in the figure. ![](/images/questions/industrial-chemistry/image-024.jpg) ## The following statements are correct ( )

  • A. A. Order of gas introduction into saturated brine: $\mathrm { NH } _ { 3 }$ then $\mathrm { CO } _ { 2 }$.
  • B. B. "Operation a" is a distillation that takes advantage of the large difference in boiling points between $\mathrm { NaHCO } _ { 3 }$ and $\mathrm { NH } _ { 4 } \mathrm { Cl }$.
  • C. C. Thermal stability: $\mathrm { NaHCO } _ { 3 } > \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$
  • D. D. The $\mathrm { Ba } ( \mathrm { OH } ) _ { 2 }$ solution can be used to identify $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ and $\mathrm { NaHCO } _ { 3 }$.

Answer: A

Solution: A. The solubility of ammonia is greater than that of carbon dioxide, in order to increase the solubility of carbon dioxide to generate more bicarbonate ions, saturated saline gas through the order: first through $\mathrm { NH } _ { 3 }$ and then through $\mathrm { CO } _ { 2 }$, so A is correct; B. "Operation a" is filtration, which utilizes the large difference in solubility between $\mathrm { NaHCO } _ { 3 }$ and $\mathrm { NH } _ { 4 } \mathrm { Cl }$, so B is wrong; C. Thermal stability: $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 } > \mathrm { NaHCO } _ { 3 }$, so C is wrong; D. $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ and $\mathrm { NaHCO } _ { 3 }$ can react with $\mathrm { Ba } ( \mathrm { OH } ) _ { 2 }$ to form a white precipitate, so you can't use $\mathrm { Ba } ( \mathrm { OH } ) _ { 2 }$ solution to identify $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ and $\mathrm { NaHCO } _ { 3 }$. INLINE_FORMULA_10]], so D is wrong; Choose A.

Question 33: 38. Chemical expert Hou Debang has made important contributions to China's soda ash industry, Hou's ...

38. Chemical expert Hou Debang has made important contributions to China's soda ash industry, Hou's alkali method of the main process shown in the figure (some substances have been omitted). The following statement is wrong ( ) ![](/images/questions/industrial-chemistry/image-025.jpg)

  • A. A. Gas $a$ is $\mathrm { NH } _ { 3 }$ and gas $b$ is $\mathrm { CO } _ { 2 }$.
  • B. B. Operation I and II are filtered
  • C. C. The ionic equation that generates $\mathrm { NaHCO } _ { 3 }$ is $\mathrm { NH } _ { 3 } + \mathrm { H } _ { 2 } \mathrm { O } + \mathrm { CO } _ { 2 } = \mathrm { HCO } _ { 3 } ^ { - } + \mathrm { NH } _ { 4 } ^ { + }$
  • D. D. Gas c and mother liquor 2 can be recycled

Answer: C

Solution: A. When preparing soda ash, in order to increase the solubility of $\mathrm { CO } _ { 2 }$, it is necessary to pass $\mathrm { NH } _ { 3 }$ and then $\mathrm { CO } _ { 2 }$, A is correct; B. Operations I and II are both processes of separating solids from liquids, so they are both filtrations, and B is correct; C. The correct ionic equation to produce $\mathrm { NaHCO } _ { 3 }$ is $\mathrm { NH } _ { 3 } + \mathrm { H } _ { 2 } \mathrm { O } + \mathrm { CO } _ { 2 } + \mathrm { Na } ^ { + } = \mathrm { NaHCO } _ { 3 } \downarrow + \mathrm { NH } _ { 4 } ^ { + }$, $\mathrm { NaHCO } _ { 3 }$ precipitates due to supersaturation and cannot be split, C is incorrect; D. From the analysis, we can see that gas c is $\mathrm { CO } _ { 2 }$ and mother liquor 2 is mainly NaCl solution, both of which can be recycled, D is correct.

Question 34: 39. Ammonia is an important chemical raw material, industrial ammonia synthesis process shown in the...

39. Ammonia is an important chemical raw material, industrial ammonia synthesis process shown in the figure below. The following analysis of the role of the regulation of production conditions is not correct ( ) ![](/images/questions/industrial-chemistry/image-026.jpg)

  • A. A. The "purification" in step (1) prevents poisoning of the catalyst.
  • B. B. Step (2) "pressurization" can speed up the reaction rate, the ammonia yield will also increase, the higher the pressure the better in industrial production
  • C. C. Step (3) is generally chosen to control the reaction temperature to $400 \sim 500 ^ { \circ } \mathrm { C }$ because the catalyst activity is maximum at this temperature
  • D. D. Step (4) can be separated by rapid cooling to liquefy the ammonia

Answer: B

Solution: A. Industrial ammonia synthesis of raw gas may contain impurities (such as sulfide, etc.), these impurities will lead to poisoning of the catalyst inactivity, step (1) "purification" can remove impurities, to prevent poisoning of the catalyst, A is correct; B. ammonia synthesis reaction for the gas volume reduction reaction, pressurization can accelerate the reaction rate and make the equilibrium positive shift to improve the ammonia yield, but the pressure is too high will significantly increase the cost of equipment and energy consumption, industrial production needs to be considered to choose the appropriate pressure (such as the title of the $2 \times 10 ^ { 7 } \mathrm {~Pa}$), is not the higher the better, B error; C. The iron catalyst used in ammonia synthesis has the greatest activity at $400 \sim 500 ^ { \circ } \mathrm { C }$, and the main purpose of controlling this temperature in step (3) is to ensure the activity of the catalyst, C is correct; D. Ammonia is easy to liquefy, while $\mathrm { N } _ { 2 } , \mathrm { H } _ { 2 }$ is difficult to liquefy, step (4) can be separated from ammonia liquefaction by rapid cooling, prompting the equilibrium to move forward, D is correct; D is correct. Therefore, B is wrong. $40 . \mathrm { C }$ [INLINE_FORMULA_3]] The writing of ionic equations, the preparation of common inorganic substances, the concept of redox reactions to determine, acids, bases, salts and oxides [Analysis]From the process given in the question, reaction 1 is the reaction of white phosphorus with excess sodium hydroxide solution to form $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ and phosphine, the chemical equation of the reaction is $\mathrm { P } _ { 4 } + 3 \mathrm { NaOH } + 3 \mathrm { H } _ { 2 } \mathrm { O } = 3 \mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 } + \mathrm { PH } _ { 3 } \uparrow$, and reaction 2 is the reaction of $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ solution with dilute sulfuric acid to form $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$, and reaction 2 is the reaction of $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ with dilute sulfuric acid to form $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$. FORMULA_7]] and sodium sulfate to form $2 \mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } = 2 \mathrm { H } _ { 3 } \mathrm { PO } _ { 2 } + \mathrm { Na } _ { 2 } \mathrm { SO } _ { 4 }$, and reaction 3 is the decomposition of $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ to form $\mathrm { H } _ { 3 } \mathrm { PO } _ { 4 }$ and $\mathrm { PH } _ { 3 }$, and the equation is $\mathrm { PH } _ { 3 }$. The chemical equation of the reaction is $2 \mathrm { H } _ { 3 } \mathrm { PO } _ { 2 } = \mathrm { H } _ { 3 } \mathrm { PO } _ { 4 } + \mathrm { PH } _ { 3 } \uparrow$, which is analyzed accordingly; A. White phosphorus and phosphine can spontaneously combust in air, so the preparation of phosphine needs to be carried out under oxygen-free conditions, A is correct; B. In reaction 1, ${ } ^ { \mathrm { P } _ { 4 } }$ produces $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ (P is +1 valence) and $\mathrm { PH } _ { 3 }$ (P is -3 valence), and the P elemental valence is increased. $( 0 \rightarrow + 1$, oxidized $)$ and reduced $( 0 \rightarrow - 3$, and reduced $)$, white phosphorus acts as an oxidizing agent and a reducing agent, B is correct; C. In reaction 1, white phosphorus reacts with excess sodium hydroxide solution to form $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ and phosphine. Reacting with sodium hydroxide solution, $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ cannot ionize hydrogen ions in the solution and belongs to the positive salt, C is wrong; D. In reaction 2, $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ (ionized $\mathrm { H } _ { 2 } \mathrm { PO } _ { 2 } ^ { - }$) reacts with dilute sulfuric acid (providing $\mathrm { H } ^ { + }$) to form $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ (a weak monoprotic acid, $\mathrm { H } ^ { + }$). difficult to ionize) with the ionic equation $\mathrm { H } _ { 2 } \mathrm { PO } _ { 2 } ^ { - } + \mathrm { H } ^ { + } = \mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ , D is correct;
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Industrial Chemistry Process

工业化工流程

34 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

Industrial Chemical Processes is an important module in the Chemistry examination, which examines students' ability to apply the principles of chemical reactions to real industrial production situations. The questions usually present the complete process of raw material treatment, separation and purification, recycling, etc., and require the analysis of material transformation, condition control and environmental benefits. These questions are highly comprehensive and need to be answered by combining knowledge of elemental compounds, reaction principles and experimental operations.

Questions:34

Key Points

  • 1Pre-treatment of raw materials and impurity removal methods in the process
  • 2Analysis of core chemical reaction principles and control of conditions
  • 3Separation and purification strategies based on solubility product and other data
  • 4Byproduct Recovery and Application of Green Chemistry Concepts
  • 5Quantitative calculations and yield/purity analysis

Study Tips

It is recommended that you read through the process to clarify the main reaction, then systematically analyze the chemistry of each step in relation to the role of reagents, the purpose of the operation, and graphical data.

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