Combustion of methane and ethane

Question

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!

Answer 1

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 : $34.84/92.96 = 0.3748 = 37.48$%