Answer
$P_{v}= 84%$
Work Step by Step
(a) The fuel is burned completely with excess air, and thus the products will contain $\mathrm{H}_2 \mathrm{O}, \mathrm{CO}_2, \mathrm{~N}_2$, and some free $\mathrm{O}_2$. Considering $1 \mathrm{kmol}$ of fuel, the combustion equation can be written as $$
\left(0.45 \mathrm{CH}_4+0.35 \mathrm{H}_2+0.20 \mathrm{~N}_2\right)+1.3 a_{\text {th }}\left(\mathrm{O}_2+3.76 \mathrm{~N}_2\right) \longrightarrow x \mathrm{CO}_2+y \mathrm{H}_2 \mathrm{O}+0.3 a_{\text {th }} \mathrm{O}_2+z \mathrm{~N}_2
$$ The unknown coefficients in the above equation are determined from mass balances, $$
\begin{aligned}
& \mathrm{C}: 0.45=x \quad \longrightarrow x=0.45 \\
& \mathrm{H}: 0.45 \times 4+0.35 \times 2=2 y \longrightarrow y=1.2 \\
& \mathrm{O}_2: 1.3 a_{\text {th }}=x+y / 2+0.3 a_{\text {th }} \longrightarrow a_{\text {th }}=1.05 \\
& \mathrm{~N}_2: 0.20+3.76 \times 1.3 a_{\text {th }}=z \longrightarrow z=5.332
\end{aligned}
$$ Thus, $$
\left(0.45 \mathrm{CH}_4+0.35 \mathrm{H}_2+0.20 \mathrm{~N}_2\right)+1.365\left(\mathrm{O}_2+3.76 \mathrm{~N}_2\right) \longrightarrow 0.45 \mathrm{CO}_2+1.2 \mathrm{H}_2 \mathrm{O}+0.315 \mathrm{O}_2+5.332 \mathrm{~N}_2
$$ The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel,
$$
\begin{aligned}
m_{\text {air }} & =(1.365 \times 4.76 \mathrm{kmol})(29 \mathrm{~kg} / \mathrm{kmol})=188.4 \mathrm{~kg} \\
m_{\text {fudl }} & =(0.45 \times 16+0.35 \times 2+0.2 \times 28) \mathrm{kg}=13.5 \mathrm{~kg}
\end{aligned}
$$ and $$
\mathrm{AF}=\frac{m_{\text {air }}}{m_{\text {fiucl }}}=\frac{188.4 \mathrm{~kg}}{13.5 \mathrm{~kg}}=13.96 \mathrm{~kg} \text { air } / \mathrm{kg} \text { fuel }
$$ (b)$$
\frac{N_v}{N_{\text {prodgas }}}=\frac{P_v}{P_{\text {prod }}} \longrightarrow \frac{1.2-N_w}{7.297-N_w}=\frac{3.1698 \mathrm{kPa}}{101.325 \mathrm{kPa}} \longrightarrow N_w=1.003\ \mathrm{kmol}
$$ since $P_{\mathrm{v}}=P_{\text {sat }} 25^{\circ} \mathrm{C}=3.1698\ \mathrm{kPa}$. Thus the fraction of water vapor that condenses is $1.003 / 1.2=0.836$ or $84 \%$.
