ChemicalCombustionmediumFE ExamCoursework

Stoichiometric Air-Fuel Ratio for Methane

Methane burns completely in dry air:

$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

Air is $21\ \text{mol}\%$ oxygen. Molar masses: $\text{CH}_4 = 16\ \text{g/mol}$ , air $= 28.97\ \text{g/mol}$ .

Find the stoichiometric air-fuel ratio on a mass basis.

Given

yO2	0.21 -	Oxygen mole fraction in air
M_CH4	16 g/mol	Methane molar mass
M_air	28.97 g/mol	Air molar mass

Hint 1

The balanced equation gives oxygen, but you are burning in air.

Hint 2

Air is only 21 mol% oxygen — scale the oxygen requirement up accordingly.

Hint 3

Then convert the molar ratio to mass using the molar masses.

Worked solution — try the problem first

Oxygen required. From the stoichiometry, 2 mol O₂ per mol CH₄.

Air required. Oxygen is only 21% of air by mole, so scale up:

$n_{air} = \frac{2}{0.21} = 9.524\ \text{mol air per mol CH}_4$

Convert to a mass basis:

$AFR = \frac{9.524 \times 28.97}{1 \times 16} = \frac{275.9}{16} = 17.24$

$\boxed{AFR \approx 17.2\ \text{kg air per kg fuel}}$

Sanity check. Stoichiometric AFR for natural gas is about 17, and for petrol about 14.7 — both well-known figures, so a result near 17 is reassuring.

Do not forget the nitrogen. Using oxygen alone would give $2(32)/16 = 4$ , which is the oxygen-fuel ratio. Nitrogen is inert but is dragged through the burner and carries away sensible heat, so it must be counted in the air mass.

Concepts:Combustion stoichiometryAir-fuel ratioTheoretical air

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