Answer
Percent theoretical air $= 144\%$
Work Step by Step
Considering $100\ \mathrm{kmol}$ of dry products, the combustion equation can be written as $$
x \mathrm{C}_8 \mathrm{H}_{18}+a \mathrm{O}_2+3.76 \mathrm{~N}_2 \longrightarrow 9.21 \mathrm{CO}_2+0.61 \mathrm{CO}+7.06 \mathrm{O}_2+83.12 \mathrm{~N}_2+b \mathrm{H}_2 \mathrm{O}
$$ The unknown coefficients $x, a$, and $b$ are determined from mass balances, $$
\begin{aligned}
& \mathrm{N}_2: 3.76 a=83.12 \longrightarrow a=22.11 \\
& \mathrm{C}: \quad 8 x=9.21+0.61 \longrightarrow x=1.23 \\
& \mathrm{H}: \quad 18 x=2 b \quad \longrightarrow b=11.07
\end{aligned}
$$ $$
\left(\text { Check O}_2: a=9.21+0.305+7.06+b / 2 \longrightarrow 22.11 \cong 22.10\right)
$$ Thus, $$
1.23 \mathrm{C}_8 \mathrm{H}_{18}+22.11\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 9.21 \mathrm{CO}_2+0.61 \mathrm{CO}+7.06 \mathrm{O}_2+83.12 \mathrm{~N}_2+11.05 \mathrm{H}_2 \mathrm{O}
$$ The combustion equation for $1 \mathrm{kmol}$ of fuel is obtained by dividing the above equation by 1.23 ,
$$
\mathrm{C}_8 \mathrm{H}_{18}+18.0\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 7.50 \mathrm{CO}_2+0.50 \mathrm{CO}+5.74 \mathrm{O}_2+67.58 \mathrm{~N}_2+9 \mathrm{H}_2 \mathrm{O}
$$ (a) The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, $$
\mathrm{AF}=\frac{m_{\text {air }}}{m_{\text {fuel }}}=\frac{(18.0 \times 4.76 \mathrm{kmol})(29 \mathrm{~kg} / \mathrm{kmol})}{(8 \mathrm{kmol})(12 \mathrm{~kg} / \mathrm{kmol})+(9 \mathrm{kmol})(2 \mathrm{~kg} / \mathrm{kmol})}=\mathbf{2 1 . 8~ k g} \text { air } / \mathrm{kg} \text { fuel }
$$ (b) To find the percent theoretical air used, we need to know the theoretical amount of air, which is determined from the theoretical combustion equation of the fuel, $$
\begin{aligned}
& \mathrm{C}_8 \mathrm{H}_{18}+a_{\text {th }} \mathrm{O}_2+3.76 \mathrm{~N}_2 \longrightarrow 8 \mathrm{CO}_2+9 \mathrm{H}_2 \mathrm{O}+3.76 a_{\text {th }} \mathrm{N}_2 \\
& \mathrm{O}_2: \quad a_{\text {th }}=8+4.5 \longrightarrow a_{\text {th }}=12.5
\end{aligned}
$$ Then, $$
\text { Percent theoretical air }=\frac{m_{\text {air,act }}}{m_{\text {air,th }}}=\frac{N_{\text {air,act }}}{N_{\text {air,th }}}=\frac{(18.0)(4.76) \mathrm{kmol}}{(12.5)(4.76) \mathrm{kmol}}=144 \%
$$
