Question

A surrogate liquid hydrocarbon fuel, approximated as \( \mathrm{C}_{10}\mathrm{H}_{12} \), is burned with dry air (79% \( \mathrm{N_2} \) and 21% \( \mathrm{O_2} \) by volume). How many moles of dry air are required for the stoichiometric combustion of 1 mole of this fuel at atmospheric conditions?

• A. 61.9
• B. 30.95
• C. 13
• D. 10

Hint:

For hydrocarbon \( \mathrm{C_xH_y} \), stoichiometric \( \mathrm{O_2} \) is \(x+y/4\) mol. Convert to air using the mole fraction of \( \mathrm{O_2} \) (0.21 for dry air).

Answer 1

The Correct Option is A

Step 1: Stoichiometric \( \mathrm{O_2} \) requirement for \( \mathrm{C_xH_y} \).
For complete combustion to \( \mathrm{CO_2} \) and \( \mathrm{H_2O} \): \[ \mathrm{C_xH_y} + \Big(x+\frac{y}{4}\Big)\mathrm{O_2} \rightarrow x\,\mathrm{CO_2} + \frac{y}{2}\,\mathrm{H_2O}. \] Here \(x = 10,\; y = 12 \;\Rightarrow\; \mathrm{O_2}\ \text{ required} = 10 + \tfrac{12}{4} = 10 + 3 = 13 \ \text{mol}\).
 

Step 2: Convert \( \mathrm{O_2} \) moles to moles of dry air.
Dry air contains 21% \( \mathrm{O_2} \) by volume (mole). \[ n_{\text{air}}=\frac{n_{\mathrm{O_2}}}{0.21}=\frac{13}{0.21}=61.9048\ldots \approx 61.9\ \text{mol}. \] 

\[\boxed{61.9\ \text{mol of dry air per mol fuel}}\]