Answer
$FA=\mathbf{0 . 0 5 7 8} \mathrm{kg} \text { fuel } / \mathrm{kg} \text { air }$
Work Step by Step
This is a theoretical combustion process since methane is burned completely with stoichiometric amount of air. The stoichiometric combustion equation of $\mathrm{CH}_4$ is $$
\mathrm{CH}_4+a_{\text {th }}\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}+3.76 a_{\text {th }} \mathrm{N}_2
$$ $ o_{2}$balance: $\quad a_{\mathrm{th}}=1+1 \quad \longrightarrow \quad a_{\mathrm{th}}=2$
Substituting,
$$
\mathrm{CH}_4+2\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}+7.52 \mathrm{~N}_2
$$ The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, $$
\mathrm{AF}=\frac{m_{\text {air }}}{m_{\text {fucl }}}=\frac{(2 \times 4.76 \mathrm{kmol})(29 \mathrm{~kg} / \mathrm{kmol})}{(1 \mathrm{kmol})(12 \mathrm{~kg} / \mathrm{kmol})+(2 \mathrm{kmol})(2 \mathrm{~kg} / \mathrm{kmol})}=17.3 \mathrm{~kg} \text { air } / \mathrm{kg} \text { fuel }
$$ The fuel-air ratio is the inverse of the air-fuel ratio, $$
\mathrm{FA}=\frac{1}{\mathrm{AF}}=\frac{1}{17.3 \mathrm{~kg} \text { air } / \mathrm{kg} \text { fuel }}=\mathbf{0 . 0 5 7 8} \mathrm{kg} \text { fuel } / \mathrm{kg} \text { air }
$$
