Ethylene oxide (C2H4O) is produced by the oxidation of ethylene (C2H4). A portio

Question

Ethylene oxide (C2H4O) is produced by the oxidation of ethylene (C2H4). A portion of the ethylene also reacts to form CO2 and H2O. The two reactions are: A reactor has two feed streams: 100 mol/s of C2H4 and 285 mol/s of air (21 mol% O2 and 79 mol% N2). The conversion of C2H4 is 90% and the yield of C2H4O is 85%. a) Based only on the first reaction, which reactant is in excess? What is the percent excess of this reactant? b) Using the Extent of Reaction method, perform a degree-of-freedom analysis to show that the molar flow rates of each component in the product stream can be determined. c) Solve for the molar flow rate of each component in the product stream (mol/s).

Explanation / Answer

Based entirely on the first reaction,

C2H4 + 0.5O2 -----> C2H4O2

a) We can say that O2 is the limiting reagent here. This means C2H4 is the reactant in excess. The C2H4 is 2 times excess then O2 in the reaction.

b) molar flow rate of C2H4 = 100 mol/s x 0.90 = 90 mol/s

molar flow rate of O2 = 285 mol/s x 0.5 x 0.21 = 30 mol/s

c) molar flow rate of C2H4O2 = 90 mol/s x 0.85 = 76.5 mol/s

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