★Specialized Examination for Chemical Engineering ★ — One question per day (2015-12-17) — Points are awarded for participation

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This post was last edited by zshiwei2008 on 2015-12-17 at 13:56. Points system for the daily question: Participation – 1~3 points; Correct answer – 5 points; In-depth analysis – 5~10 points. The answer is provided the next day. Due to limited capabilities, I am unable to provide more in-depth explanations for each question; therefore, on the second day, in addition to offering reference options, I will also remove the restriction on reading replies. I hope that all candidates can learn from one another and improve together! Daily Question (2015-12-17): 1. Prepare 100 kg of 80% sulfuric acid using 98% concentrated sulfuric acid and 40% dilute sulfuric acid. Then the amount of raw material acid required is ( ). A. A large amount of concentrated sulfuric acid is used. B. A large amount of dilute sulfuric acid is used. C. Half concentrated sulfuric acid and half dilute sulfuric acid are used. D. It is not possible to prepare it. 2. If the conversion rate of the key reactant in the reaction is 85% and the selectivity is 78%, what is the yield of the desired product? A. 66.3% B. 91.76% C. 85% D. 78% Answer: 1 A A. The answer is correct. Since the concentration of sulfuric acid obtained in the final mixture is close to that of the raw acid, a large amount of concentrated sulfuric acid is required. Alternatively, the following material balance calculation can be performed: Let x kg be the amount of 98% concentrated sulfuric acid required. Let y kg be the amount of 40% dilute sulfuric acid required. Then we have x + y = 100 and 0.98x + 0.4y = 80. Solving these equations gives x = 68.97 kg and y = 31.03 kg. It can also be solved using the following method: 0.98x + 0.4y = 0.8(x + y); 0.98x – 0.8x = 0.8y – 0.4y; 0.18x = 0.4y; x/y = 0.4/0.18 = 2.22. A. The answer is correct; B, C, and D are incorrect. The yield Y of the product = X·S, that is, yield = conversion rate × selectivity.