Answer
Theoratical air = $186\%$
Work Step by Step
( $a$ ) The combustion equation in this case can be written as $$
\mathrm{C}_2 \mathrm{H}_6+a\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}+3 \mathrm{O}_2+3.76 a \mathrm{~N}_2
$$ $\mathrm{O}_2$ balance: $$
\mathrm{a}=2+1.5+3 \longrightarrow \mathrm{a}=6.5 $$ Substituting, $$
\mathrm{C}_2 \mathrm{H}_6+6.5\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}+3 \mathrm{O}_2+24.44 \mathrm{~N}_2
$$ The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, $$
\mathrm{AF}=\frac{m_{\text {air }}}{m_{\text {fucl }}}=\frac{(6.5 \times 4.76 \mathrm{kmol})(29 \mathrm{~kg} / \mathrm{kmol})}{(2 \mathrm{kmol})(12 \mathrm{~kg} / \mathrm{kmol})+(3 \mathrm{kmol})(2 \mathrm{~kg} / \mathrm{kmol})}=\mathbf{2 9 . 9}\ \mathrm{kg} \text { air } / \mathrm{kg} \text { fuel }
$$ (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 $\mathrm{C}_2 \mathrm{H}_6$,
$$
\mathrm{C}_2 \mathrm{H}_6+a_{\text {th }} \mathrm{O}_2+3.76 \mathrm{~N}_2 \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}+3.76 a_{\text {th }} \mathrm{N}_2
$$ $\mathrm{O}_2$ balance:
$\quad a_{\mathrm{th}}=2+1.5 \longrightarrow a_{\mathrm{th}}=3.5$
Then, $$
\text { Percent theoretical air }=\frac{m_{\text {air,act }}}{m_{\text {air,th }}}=\frac{N_{\text {air,act }}}{N_{\text {air,th }}}=\frac{a}{a_{\text {th }}}=\frac{6.5}{3.5}\\=186 \%
$$
