1) Consider a 100 mol mixture that is 82.0% methane (CH4) and 18.0% ethane (C2H6

Question

1) Consider a 100 mol mixture that is 82.0% methane (CH4) and 18.0% ethane (C2H6). To this mixture is added 26.0% excess air. Of the methane present, 93.20% reacts, 91.80% of which forms carbon dioxide (CO2), and the balance forms carbon monoxide (CO). Of the ethane present, 85.2% reacts, 91.80% of which forms carbon dioxide, and the balance forms carbon monoxide. What is the theoretical amount of oxygen required for the fuel mixture?

a) What is the theoretical amount of oxygen required for the fuel mixture? ___ mol O2

b) What amount of air is added to the fuel mixture? ___ mol air

c) How many moles of methane are present in the product gas? ___ mol CH4

d) How many moles of ethane are present in the product gas? ___ mol C2H6

e) How many moles of carbon dioxide are present in the product gas? ___ mol CO2

f) How many moles of carbon monoxide are present in the product gas? ___ mol CO

g) How many moles of water vapor are present in the product gas? ___ mol H2O

h) How many moles of oxygen are present in the product gas? ___ mol O2

i) How many moles of nitrgen are present in the product gas? ___ mol N2

Explanation / Answer

100 mol mixture, 82.0% methane (CH4) and 18.0% ethane (C2H6)

Therefore, 82 mol CH4 and 18 mol C2H6

Total Methane consumed in reaction with oxygen = 93.2/100 * 82 = 76.42 mol

Methane consumed in forming CO2 = 91.8/100 * 76.42 = 70.15 mol

Methane consumed in forming CO = 76.42 -70.15 = 6.27 mol

Total Ethane consumed in reaction with oxygen = 85.2/100 * 18 = 15.33 mol

Ethane consumed in forming CO2 = 91.8/100 * 15.33 = 14.07 mol

Ethane consumed in forming CO = 15.33 – 14.07 = 1.26 mol

Reactions,

CH4 + O2            CO2 + 2H2O

2CH4 + 3O2         2CO + 4H2O

2C2H6 + 7O2         4CO2 + 6H2O

2C2H6 + 5O2         4CO + 6H2O

a) The theoretical amount of oxygen required for the fuel mixture is 70.5 + 9.4 + 51.4 + 3.15 = 134.45 mol O2

b) The amount of air added to the fuel mixture is 20.95/100 * x = 134.45

therefore x = 641.7 mol of air is required (as 20.95% oxygen is present in air) but since 26% excess air is taken, the amount of air added is (26/100*641.7 + 641.7) = 808.85 mol of air

c) No. of moles of methane present in the product gas is (82-76.42) = 5.58 mol CH4

d) No. of moles of ethane present in the product gas is (18-15.33)= 2.67 mol C2H6

e) No. of moles of carbon dioxide present in the product gas is (70.1+14.07*2) = 98.29 mol CO2

f) No. of moles of carbon monoxide present in the product gas is (6.27+1.26*2) = 8.79 mol CO

g) No. of moles of water vapor present in the product gas (70.15*2+6.27*2+14.07*3+1.26*3) = (140.3 + 12.54 + 42.21 + 3.78) = 198.8 mol H2O

h) No. of moles of oxygen present in the product gas is (20.95/100*808.85 - 134.45) = 35 mol O2

i) No. of moles of nitrogen present in the product gas is 78.09/100*808.85 = 630.9 mol N2 (as 78.09% nitrogen is present in air)

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