please help me and show full answer. Process engineering Ethylene oxide (C2H40)

Question

please help me and show full answer.
Process engineering Ethylene oxide (C2H40) is produced by the oxidation of ethylene C2H). A portion o ethylene also reacts to form CO2 and H2O·The two reactions are: tion of ethylene (C2Hs). A portion of the CH4 + 0.5 O2 C2H40 C2H4 + 302 2 CO2 + 2 H2O A reactor has two feed streams: 100 mols of CH4 and 285 mol/s of air (21 mol% 02 and 79 mol% N) The conversion of C2H4 is 90% and the yield of C2H4 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 c) Solve for the molar flow rate of each component in the product stream (mol/s). the molar flow rates of each component in the product stream can be determined.

Explanation / Answer

The reactions are

C2H4 + 1/2 O2 --> C2H4O

C2H4 + 3O2 --> 2CO2 + 2H2O

1) the flow rate of ethylene = 100 mol /second

the flow rate of air = 285 mol / second

in each 100 mol of air it contains 21 mol of oxygen

so in 285 moles of air it containse 59.85 moles of oxygen

So ethylene is the limiting reagent for first reaction and oxygen is the excess reagent

Fo 1 mole of ethylene oxidation 0.5 moles of oxygen is required so

100 moles will require 50moles of oxygen

The excess of oxygen = 59.85 - 50 = 9.85 moles

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