Answer to Question #116432 in General Chemistry for Xean

Question

100kg fuel is known to consist of 8% CO2, 12% CO, 44% CH4, and 36% C8H16. The air used

to burn this amount of fuel was supplied with 45% excess of the theoretical amount. Determine
the complete composition of the combustion gases.

Answer 1

100kg fuel consists of 8 kg CO2, 12 kg CO, 44 kg CH4, and 36 kg C8H16

8 kg CO2 goes to the products with no change

M (CO) = 28.01

M (CO2) = 44.01

M (O2) = 32.00

M (CH4) = 16.04

M (C8H16) = 112.21

2CO + O2 = 2CO2

12 kg * 44.01/28.01 = 18.85 kg of CO2 produced

12 kg * 32.00/2*20.01 = 6.85 kg O2 used

CH4 + 2O2 = CO2 + 2H2O

44 kg * 44.01/16.04 = 120.7 kg CO2 produced

44*2*32.00/16.04 = 175.56 kg O2 used

C8H16 + 12O2 = 8CO2 + 8H2O

36*8*44.01/112.21 = 112.96 kg CO2 produced

36*12*32.00/112.21 = 123.2 kg O2 used

Total CO2:

8 + 18.85 + 120.7 + 112.96 = 260.51 kg

Total O2:

6.85 + 175.56 + 123.2 = 305.61 kg

Air is a mixture of several gases, where the two most dominant components in dry air are 20% oxygen and 78% nitrogen 1% - other gases.

The mass of air used:

305.61 / 0.20 = 1528.05 kg

+ 45% of excess: 1528.05 + 687.6 = 2215.65 kg

It consists of 78% of N2: 0.78 * 2215.65 kg = 1728.2 kg (products part)

So products:

260.51 kg of CO2

1728.2 kg N2 (which does not react)

Assumption: water is at liquid state.