Answer
$9.42 \text{ kg air /kg fuel}$
Work Step by Step
Considering $1\ \mathrm{kmol}$ of fuel, the combustion equation can be written as $$
\left(0.65 \mathrm{CH}_4+0.08 \mathrm{H}_2+0.18 \mathrm{~N}_2+0.03 \mathrm{O}_2+0.06 \mathrm{CO}_2\right)+a_{\text {th }}\left(\mathrm{O}_2+3.76 \mathrm{~N}_2\right) \longrightarrow x \mathrm{CO}_2+y \mathrm{H}_2 \mathrm{O}+z \mathrm{~N}_2
$$ The unknown coefficients in the above equation are determined from mass balances, $$
\begin{aligned}
\mathrm{C}:& 0.65+0.06=x \quad \longrightarrow x=0.71 \\
\mathrm{H}:& 0.65 \times 4+0.08 \times 2=2 y \longrightarrow y=1.38 \\
\mathrm{O}_2:& 03+0.06+a_{\text {th }}=x+y / 2 \longrightarrow a_{\text {th }}=1.31 \\
\mathrm{~N}_2:& 0.18+3.76 a_{\text {th }}=z \quad \longrightarrow z=5.106
\end{aligned}
$$ Thus, $$
\begin{aligned}
&\left(0.65 \mathrm{CH}_4+0.08 \mathrm{H}_2+0.18 \mathrm{~N}_2+0.03 \mathrm{O}_2+0.06 \mathrm{CO}_2\right)+1.31\left(\mathrm{O}_2+3.76 \mathrm{~N}_2\right) \\
& \longrightarrow 0.71 \mathrm{CO}_2+1.38 \mathrm{H}_2 \mathrm{O}+5.106 \mathrm{~N}_2
\end{aligned}
$$ The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel, $
\begin{aligned}
m_{\text {air }} & =(1.31 \times 4.76 \mathrm{kmol})(29 \mathrm{~kg} / \mathrm{kmol})=180.8 \mathrm{~kg} \\
m_{\text {fud }} & =(0.65 \times 16+0.08 \times 2+0.18 \times 28+0.03 \times 32+0.06 \times 44) \mathrm{kg}=19.2 \mathrm{~kg}
\end{aligned}
$ and $$
\mathrm{AF}_{\text {th }}=\frac{m_{\text {air,th }}}{m_{\text {fucl }}}=\frac{180.8 \mathrm{~kg}}{19.2 \mathrm{~kg}}=9.42 \mathrm{~kg} \text { air } / \mathrm{kg} \text { fuel }
$$
