1 question on the 11-year Chemical Engineering case study
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This post was last edited by forrestwong on 2016-7-15 12:16. 1. A mixture of 1 kg of NaCl and KCl was treated with H2SO4, resulting in the recovery of 1.2 kg of K2SO4 and Na2SO4. What is the weight ratio of NaCl to KCl in the original mixture? The known reaction is: NaCl + KCl + H2SO4 → 1/2Na2SO4 + 1/2K2SO4 + HCl. (A) 0.2 (B) 2.7 (C) 0.5 (D) 2.1. It’s not clear in this question which reagent will be in excess and which one will not, and there are no given conditions to help determine this. Based on the assumption that an excess amount is considered, the mass ratios come out to be 0.77 and 0.80 respectively, and the results simply don’t match at all. Also, the reference answer was calculated without using the reaction stoichiometric coefficients. What’s going on? Please let fellow sailors discuss. Thank you.
@zhanghp30 @Higee
Solution: Let the amount of NaCl be x moles and that of KCl be y moles. The molecular weight of NaCl is 58.5; that of KCl is 74.5; the molecular weight of Na2SO4 is 142; and that of K2SO4 is 174. Applying the law of conservation of elements gives us: 58.5x + 74.5y = 1000 and 142x/2 + 174y/2 = 12000. From these equations, we obtain x = 12 moles and y = 4 moles. Additionally, 58.5x/74.5y = 702/298 ≈ 2.356, so the correct answer is D
The meaning is that there is an excess of sulfuric acid, and the answer key is correct.
The question is how do you know whether it’s an excess of NaCl or an excess of KCl.
This post was last edited by forrestwong on 2016-7-15 17:25. NaCl+KCl+ H2SO4→1/2Na2SO4+1/2K2SO4+2HCl; 2NaCl+H2SO4→Na2SO4+2HCl; 2KCl+ H2SO4→K2SO4+2HCl. When one of these substances is in excess, one of the above chemical reactions takes place. Then, by element conservation: X*58.5 + Y*74.5 = 100. Also, X/2*142 + Y/2*174 = 1200 (Y is greater than X), or X/2*142 + Y/2*174 = 1200 (X is greater than Y); the results of these two equations are the same. Then solve the system together. I’m not sure if what I wrote is correct
2NaCl + H2SO4 → Na2SO4 + 2HCl 2KCl + H2SO4 → K2SO4 + 2HCl. In this reaction, there is no issue of which substance, NaCl or KCl, is in excess. Are the two equations you wrote above different from NaCl + KCl + H2SO4 → 1/2Na2SO4 + 1/2K2SO4 + 2HCl? The equations written earlier, 58.5x+74.5y=1000 and 142x/2+174y/2=1200, are incorrect
This question provides the following known reaction equation: NaCl + KCl + H2SO4 → 1/2Na2SO4 + 1/2K2SO4 + HCl. It’s quite misleading; it’s actually two separate reaction equations, yet they are presented as one, leading to the misconception that the ratio of NaCl to KCl is 1:1, when in fact it isn’t.
There are two reasons why one can’t solve this problem: one is that the thinking gets stuck in a dead end, and the other is that the person is completely unqualified for it. If one gets stuck in a dead end, then they just need to find a way out of it. It’s a matter of conservation of mass – it doesn’t matter whether something is in excess or not. . .