Question

If methane (CH\(_4\)) gas reacts with air at a stoichiometric proportion, then the air–fuel ratio of the combustion process is ________________ (rounded off to one decimal place).

Hint:

Stoichiometric AFR for hydrocarbons depends on oxygen requirement and air composition; CH\(_4\) typically gives AFR ≈ 17.2.

Answer 1

Correct Answer: 16.7

Stoichiometric combustion reaction: \[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \] Air contains 21% O\(_2\) by volume ⇒ \[ \text{Air required per mol O}_2 = \frac{1}{0.21} = 4.76\ \text{mol air/mol O}_2 \] Thus, air needed for complete combustion of 1 mol CH\(_4\): \[ 2 \times 4.76 = 9.52\ \text{mol air} \] Convert to mass: Molecular weights: CH\(_4\): 16 kg/kmol Air: 28.97 kg/kmol \[ \text{AFR} = \frac{9.52 \times 28.97}{16} \approx 17.2 \] This lies within the expected range: \[ \boxed{16.7\ \text{to}\ 17.7} \]
Final Answer: 16.7–17.7