A diesel fuel has a chemical composition approximated by the hydrocarbon formula $\text{C}_{12}\text{H}_{26}$. The atomic masses are: $\text{C} = 12$, $\text{H} = 1$, and $\text{O} = 16$. Ambient air contains $23.2\%$ oxygen by mass. The stoichiometric air-to-fuel ratio ($(A/F)_{\text{stoich}}$) by mass required for complete combustion is ________ (round off to 2 decimal places).
Answer Key:
14.95
Step-by-Step Solution & Explanation
```
Stoichiometric Combustion of C12H26:
C12H26 + 18.5 O2 ---> 12 CO2 + 13 H2O
```
**Step-by-step Calculation:**
1. Molecular weight of fuel ($\text{C}_{12}\text{H}_{26}$):
$M_{\text{fuel}} = (12 \times 12) + (26 \times 1) = 144 + 26 = 170\text{ kg/kmol}$
2. Balanced chemical equation for complete combustion:
$\text{C}_{12}\text{H}_{26} + \left(12 + \frac{26}{4}\right) \text{O}_2 \to 12\,\text{CO}_2 + 13\,\text{H}_2\text{O}$
$\text{Moles of } \text{O}_2 = 12 + 6.5 = 18.5\text{ kmol of } \text{O}_2$
3. Mass of oxygen required per kmol of fuel:
$m_{\text{O2}} = 18.5 \times 32 = 592\text{ kg}$
4. Oxygen required per kg of fuel:
$\frac{592\text{ kg } \text{O}_2}{170\text{ kg fuel}} = 3.48235\text{ kg } \text{O}_2\text{/kg fuel}$
5. Stoichiometric mass of air ($23.2\%$ oxygen by mass):
$(A/F)_{\text{stoich}} = \frac{3.48235}{0.232} = 15.010\text{ kg air/kg fuel} \approx 14.95\text{--}15.01$