Calculate the stoichiometric (theoretical) mass of air required for the complete combustion of $1.0\text{ kg}$ of diesel fuel assumed to have chemical formula $\text{C}_{12}\text{H}_{26}$ (atomic weights: $\text{C}=12$, $\text{H}=1$, $\text{O}=16$; air contains $23.2\%$ oxygen by mass; round off to 2 decimal places).
Answer Key:
14.97
Step-by-Step Solution & Explanation
Combustion reaction:
$\text{C}_{12}\text{H}_{26} + \left(12 + \frac{26}{4}\right)\text{O}_2 \rightarrow 12\text{CO}_2 + 13\text{H}_2\text{O}$
Moles of $\text{O}_2 = 12 + 6.5 = 18.5\text{ moles}$.
Molecular weight of $\text{C}_{12}\text{H}_{26} = (12 \times 12) + (26 \times 1) = 144 + 26 = 170\text{ g/mol}$.
Mass of $\text{O}_2$ required per mole of fuel $= 18.5 \times 32 = 592\text{ g}$.
Mass of $\text{O}_2$ per kg of fuel:
$m_{\text{O}_2} = \frac{592}{170} \approx 3.48235\text{ kg }\text{O}_2$
Stoichiometric air required (air has $23.2\%$ $\text{O}_2$ by mass):
$m_{\text{air}} = \frac{3.48235}{0.232} \approx 14.97\text{ kg air/kg fuel}$
Acceptable range: $14.8 - 15.2$.