# Combustion of methane and ethane

I'm having trouble with the following problem:

A gas fuel of Methane ($$\ce{CH4}$$) and Ethane ($$\ce{C2H6}$$) is burned with atmospheric air, and the volumetric analysis of the products of combustion yields the following: 5.32% $$\ce{CO2}$$, 1.60% $$\ce{CO}$$, 7.32% $$\ce{O2}$$, 12.24% $$\ce{H2O}$$ and the rest $$\ce{N2}$$.

Calculate the percentage of excess air and the percentage of Methane and Ethane in the fuel.

The balance equation from the data given above should be:

$$x \ce{CH4}+y \ce{C2H6} +a(\ce{O2}+3.76 \ce{N2})\rightarrow 5.32\ce{CO2}+ 1.60\ce{CO}+7.32 \ce{O2}+12.24 \ce{H2O}+73.52\ce{N2}$$

Balancing each element:

$$\ce{N2}: \quad 3.76 a =73.52 \\ \ce{C}: \quad x+2y=5.32+1.60 \\ \ce{H}: \quad 4x+6y=24.48$$

And solving the system I get: $$a=19.55, \quad x=3.72, \quad y=1.60$$

How can I get the percentage of excess air and types of fuel from there? Thanks!

• You can get the fuel composition from the total H/C ratio of products. – Poutnik Oct 17 '20 at 18:57

The excess of $$\ce{O2}$$ is $${7.32~ mol}$$. So the excess air is : $${7.32 ~mol· (1 + 3.76) = 34.84~ mol}$$.
The total amount of air is : $${a·4.76 = 19.53~ mol ·4.76 = 92.96~ mol}$$.
The percentage of excess air is : $$34.84/92.96 = 0.3748 = 37.48$$%