Tiantai University version: Professional practice questions 1-7 – any questions?

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1-7 In the reaction for producing ethanol from ethylene hydration, the main and side reactions are as follows: Main reaction: C2H4 + H2O → C2H5OH; Side reaction: 2C2H5OH → (C2H5)2O + H2O. If the reactor feed consists of 53.7% C2H4 (by mole percentage, the same hereafter), 36.7% H2O, and the remainder being inert gases. The conversion rate of ethylene is 5%, and the yield of ethanol is 90%, based on the consumption of ethylene. Then the total molar amount of the material output from the reactor is ( ). A. 87.72 mol B. 87.98 mol C. 88.12 mol D. 87.58 mol Answer: 1-7 A A. The answer is correct. The calculation basis is set at 100 mol of feedstock, and a material balance is conducted for the process. The amount of ethylene used is 53.7 mol; the number of moles of ethanol produced is 53.7×0.05×0.9 = 2.42 mol. The number of moles of diethyl ether produced is equal to half of the difference between the theoretical number of moles of ethanol that could be formed from the ethylene and the actual amount of ethanol obtained. Because according to the reaction equation, 1 mol of diethyl ether is produced from every 2 mol of ethanol. Therefore, the amount of diethyl ether produced is (53.7×0.05–2.42)/2 = 0.133 mol. The number of moles of water used is 36.7 mol; the number of moles of water that participates in the reaction is 36.7–53.7×0.05 = 36.7–2.685 = 34.015 mol. The amount of water generated as a by-product is 0.133 mol. Thus, the total number of moles of water is 33.815+0.133 = 34.15 mol. The composition of the products is as follows: ethylene – 53.7×(1–0.05) = 51.02 mol, accounting for 58.26%; ethanol – 2.42 mol, accounting for 2.76%; diethyl ether – 0.133 mol, accounting for 0.15%. Water – 34.15 mol, accounting for 38.8%. The total number of moles of all products is 87.72 mol. Question: Why isn’t the amount of inert gas included?