Answer
$P_{v}=3.1698\text{ kPa}\\y_{v}= 0.1652\\P_{min}=19.2\text{ kPa}$
Work Step by Step
The stoichiometric equation for this reaction is $$
\mathrm{C}_2 \mathrm{H}_6+3.5\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}+13.16 \mathrm{~N}_2
$$ At the minimum pressure, the product mixture will be saturated with water vapor and $$
P_v=P_{\text {sat } @ 25^{\circ} \mathrm{C}}=3.1698\ \mathrm{kPa}
$$ The mole fraction of water in the products is $$
y_v=\frac{N_{\mathrm{H} 2 \mathrm{O}}}{N_{\text {prod }}}=\frac{3 \mathrm{kmol}}{(2+3+13.16) \mathrm{kmol}}=0.1652
$$ The minimum pressure of the products is then $$
P_{\min }=\frac{P_v}{y_v}=\frac{3.1698 \mathrm{kPa}}{0.1652}=19.2\ \mathrm{kPa}
$$
