Answer
(a) 5.3
Work Step by Step
Propane ($C_3H_8$) is burned with $150 \%$ theoretical air.
1. Molar mass of propane ($C_3H_8$): $44.1 \mathrm{~g} / \mathrm{mol}$.
2. Mass of air required for complete combustion of 1 mole of propane: $160 \mathrm{~g} / \mathrm{mol}$.
3. Mass of air required for $150 \%$ theoretical air (excess air): $240 \mathrm{~g} / \mathrm{mol}$.
4. Air-fuel mass ratio:
Air-Fuel Mass Ratio $=$ Mass of Air with excess air $/$ Molar Mass of Propane
$$
\begin{aligned}
& =240 \mathrm{~g} / \mathrm{mol} / 44.1 \mathrm{~g} / \mathrm{mol} \\
& \approx 5.45 .
\end{aligned}
$$Rounded to one decimal place, the closest option is (a) $5.3$.
