Consider a 100 mol mixture that is 70.0% methane (CH4) and 30.0% ethane (C2H6).

Question

Consider a 100 mol mixture that is 70.0% methane (CH4) and 30.0% ethane (C2H6). To this mixture is added 27.0% excess air. Of the methane present, 92.40% reacts, 89.60% of which forms carbon dioxide (CO2), and the balance forms carbon monoxide (CO). Of the ethane present, 88.5% reacts, 89.60% of which forms carbon dioxide, and the balance forms carbon monoxide. What is the theoretical amount of oxygen required for the fuel mixture?

What amount of air is added to the fuel mixture?

How many moles of methane are present in the product gas?

How many moles of ethane are present in the product gas?

How many moles of carbon dioxide are present in the product gas?

How many moles of carbon monoxide are present in the product gas?

How many moles of water vapor are present in the product gas?

How many moles of oxygen are present in the product gas?

Explanation / Answer

total moles of gas mixture = 100

moles of methane = 70 & moles of ethane = 30

Now, CH4 + 2O2 -------> CO2 + 2H2O

CH4 + (3/2)O2 -------> CO + 2H2O

Also,

C2H6 + (7/2)O2 -------> 2CO2 + 3H2O

C2H6 + (5/2)O2 -------> 2CO + 3H2O

Moles of methane unreacted = 70 - (0.924*70) = 5.32 = moles of methane present in the product gas

moles of ethane unreacted = 30 - (0.889*30) = 3.33 = moles of ethane present in the product gas

Thus, moles of methane reacted = 70-5.32 = 64.68

moles of methane forming CO2 = 0.896*64.68 = 57.95

Thus, moles of CO2 formed = moles of methane reacting = 57.95

moles of oxygen required = 2*moles of methane forming CO2 = 115.9

moles of methane forming CO = 64.68-57.95 = 6.73

moles of CO formed by methane = moles of methane reacting = 6.73

moles of O2 required = (3/2)*moles of methane reacting = 86.925

Now,

moles of ethane reacted = 30-3.33 = 26.67

moles of ethane forming CO2 = 0.896*26.67 = 23.9

Thus, moles of CO2 formed = 2*moles of ethane reacting = 47.8

moles of oxygen required = (7/2)*moles of ethane forming CO2 = 83.65

moles of ethane forming CO = 26.67 - 23.9 = 2.77

moles of CO formed by ethane = 2*moles of ethane reacting = 5.54

moles of O2 required = (5/2)*moles of ethane reacting = 6.925

Thus, total moles of CO2 present in the product gas = 57.95 + 47.8 = 105.75

moles of CO formed = 6.73+5.54 = 12.27

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