Answer
$AF_{}= 12.94\text{ kg air/kg fuel fuel}$
Work Step by Step
Analysis The balanced reaction equation for stoichiometric air is $$
\mathrm{CH}_3 \mathrm{OH}+a_{\text {th }}\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}+a_{\text {th }} \times 3.76 \mathrm{~N}_2
$$ The stoicihiometric coefficient $a_{\mathrm{th}}$ is determined from an $\mathrm{O}_2$ balance: $$
0.5+a_{\text {th }}=1+1 \longrightarrow a_{\text {th }}=1.5
$$ Substituting, $$
\mathrm{CH}_3 \mathrm{OH}+1.5\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}+1.5 \times 3.76 \mathrm{~N}_2
$$ The reaction with $100 \%$ excess air and incomplete combustion can be written as $$
\mathrm{CH}_3 \mathrm{OH}+2 \times 1.5\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 0.60 \mathrm{CO}_2+0.40 \mathrm{CO}+2 \mathrm{H}_2 \mathrm{O}+x \mathrm{O}_2+2 \times 1.5 \times 3.76 \mathrm{~N}_2
$$ The coefficient for $\mathrm{O}_2$ is determined from a mass balance,
$\mathrm{O}_2$ balance: $$
0.5+2 \times 1.5=0.6+0.2+1+x \longrightarrow x=1.7
$$ Substituting, $$
\mathrm{CH}_3 \mathrm{OH}+3\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 0.6 \mathrm{CO}_2+0.4 \mathrm{CO}+2 \mathrm{H}_2 \mathrm{O}+1.7 \mathrm{O}_2+11.28 \mathrm{~N}_2
$$ The air-fuel mass ratio is $$
\mathrm{AF}=\frac{m_{\text {air }}}{m_{\text {fuel }}}=\frac{(3 \times 4.76 \times 29) \mathrm{kg}}{(1 \times 32) \mathrm{kg}}=\frac{414.1 \mathrm{~kg}}{32 \mathrm{~kg}}=12.94\ \mathrm{~kg} \text { air } / \mathrm{kg} \text { fuel }
$$
