Answer
$D=a n$
$E=b n$
$F=m / 2$
$J=3.76(1+B) A_{\text {ih }}$
$G=\frac{b n}{2}+B A_{\text {th }}$
Work Step by Step
The balanced reaction equation for stoichiometric air is $$
\mathrm{C}_n \mathrm{H}_m+A_{\mathrm{th}}\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow n \mathrm{CO}_2+(\mathrm{m} / 2) \mathrm{H}_2 \mathrm{O}+3.76 A_{\mathrm{lh}} \mathrm{N}_2
$$ The stoichiometric coefficient $A_{\text {出 }}$ is determined from an $\mathrm{O}_2$ balance: $$
A_{\mathrm{th}}=n+m / 4
$$ The reaction with excess air and incomplete combustion is $$
\mathrm{C}_n \mathrm{H}_m+(\mathrm{I}+B) A_{\mathrm{lh}}\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow a n \mathrm{CO}_2+b n \mathrm{CO}+(m / 2) \mathrm{H}_2 \mathrm{O}+G \mathrm{O}_2+3.76(\mathrm{I}+B) A_{\mathrm{th}} \mathrm{N}_2
$$ The given reaction is $$
\mathrm{C}_n \mathrm{H}_n+(\mathrm{I}+B) A_{\mathrm{th}}\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow D \mathrm{CO}_2+E \mathrm{CO}+F \mathrm{H}_2 \mathrm{O}+G \mathrm{O}_2+J \mathrm{~N}_2
$$ Thus, $$
\begin{aligned}
& D=a n \\
& E=b n \\
& F=m / 2 \\
& J=3.76(1+B) A_{\text {ih }}
\end{aligned}
$$ The coefficient $G$ for $\mathrm{O}_2$ is determined from a mass balance, $\mathrm{O}_2$ balance:$$
\begin{aligned}
(1+B) A_{\text {th }} & =a n+\frac{b n}{2}+\frac{m}{4}+G \\
(1+B)\left(n+\frac{m}{4}\right) & =a n+\frac{b n}{2}+\frac{m}{4}+G \\
\left(n+\frac{m}{4}\right)+B A_{\text {lh }} & =a n+\frac{b n}{2}+\frac{m}{4}+G \\
G & =n+B A_{\text {th }}-a n-\frac{b n}{2} \\
& =n(1-a)+B A_{\text {th }}-\frac{b n}{2} \\
& =n b-\frac{b n}{2}+B A_{\text {Lh }} \\
& =\frac{b n}{2}+B A_{\text {th }}
\end{aligned}
$$
