Answer
$37\%$
Work Step by Step
The theoretical combustion equation in this case can be written as
$$
\mathrm{C}_2 \mathrm{H}_6+a_{\text {th }} \mathrm{O}_2+3.76 \mathrm{~N}_2 \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}+3.76 a_{\text {th }} \mathrm{N}_2
$$ where $a_{\text { }}$ is the stoichiometric coefficient for air. It is determined from $O_2$ balance: $\quad a_{\text {th }}=2+1.5 \longrightarrow a_{\text {th }}=3.5$
The air-fuel ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the mass of the fuel for, $$
\mathrm{AF}_{\text {th }}=\frac{m_{\text {air, th }}}{m_{\text {fuel }}}=\frac{(3.5 \times 4.76 \mathrm{kmol})(29 \mathrm{~kg} / \mathrm{kmol})}{(2 \mathrm{kmol})(12 \mathrm{~kg} / \mathrm{kmol})+(3 \mathrm{kmol})(2 \mathrm{~kg} / \mathrm{kmol})}=16.1 \mathrm{~kg} \text { air } / \mathrm{kg}
$$ The actual air-fuel ratio used is $$
\mathrm{AF}_{\text {act }}=\frac{\dot{m}_{\text {uir }}}{\dot{m}_{\text {fuel }}}=\frac{176 \mathrm{~kg} / \mathrm{h}}{8 \mathrm{~kg} / \mathrm{h}}=22 \mathrm{kgair} / \mathrm{kgfuel}
$$ Then the percent theoretical air used can be determined from $$
\text { Percent theoretical air }=\frac{\mathrm{AF}_{\text {act }}}{\mathrm{AF}_{\text {th }}}=\frac{22 \mathrm{~kg} \text { air } / \mathrm{kg} \text { fuel }}{16.1 \mathrm{~kg} \text { air } / \mathrm{kg} \text { fuel }}=137 \%
$$ Thus the excess air used during this process is $\mathbf{3 7 \%}$.
