Answer
$y_{}= 0.1550$
$P_{v}= 1.6951\text{ psia}$
$y_{g}=0.1153$
$f_{vapor}=0.744$
Work Step by Step
The fuel is burned completely with the air, and thus the products will contain only $\mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}$, and $\mathrm{N}_2$. Considering $1 \mathrm{kmol} \mathrm{C}_3 \mathrm{H}_8$, the combustion equation can be written as$$
\mathrm{C}_3 \mathrm{H}_8+5\left(\mathrm{O}_2+3.76 \mathrm{~N}_2\right) \longrightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}+18.8 \mathrm{~N}_2 $$
The mole fraction of water in the products is $$
y=\frac{N_{\mathrm{H} 2 \mathrm{O}}}{N_{\mathrm{pmod}}}=\frac{4 \mathrm{kmol}}{(3+4+18.8) \mathrm{kmol}}=0.1550
$$ The saturation pressure for the water vapor is $$
P_v=P_{\text {sat } @ 120 \% \mathrm{~F}}=1.6951\ \mathrm{psia}
$$ When the combustion gases are saturated, the mole fraction of the water vapor will be $$
y_g=\frac{P_v}{P}=\frac{1.6951 \mathrm{kPa}}{14.696 \mathrm{kPa}}=0.1153
$$ Thus, the fraction of water vapor in the combustion products is $$
f_{\text {vapor }}=\frac{y_g}{y}=\frac{0.1153}{0.1550}=\mathbf{0 . 7 4 4}
$$
