Combustion of octane with excess air

Stoichiometric reaction

$$\text{C}_8\text{H}_{18} + 12.5\,(\text{O}_2 + 3.76\,\text{N}_2) \rightarrow 8\,\text{CO}_2 + 9\,\text{H}_2\text{O} + 47\,\text{N}_2$$

Oxygen needed: 8 (for C) + 4.5 (for H) = 12.5 mol per mol of fuel.

(a) Air-fuel ratio

$$\begin{aligned} (A/F)_{stoich} &= \frac{12.5 \times 4.76 \times 28.97}{114.23} = \frac{1723.7}{114.23} = 15.09 \\ (A/F)_{actual} &= 1.20 \times 15.09 = 18.11 \end{aligned}$$

Reaction with 120% theoretical air

Oxygen supplied $= 1.2 \times 12.5 = 15$ mol, nitrogen $= 15 \times 3.76 = 56.4$ mol, excess O$_2$ $= 15 - 12.5 = 2.5$ mol.

$$\text{C}_8\text{H}_{18} + 15\,(\text{O}_2 + 3.76\,\text{N}_2) \rightarrow 8\,\text{CO}_2 + 9\,\text{H}_2\text{O} + 2.5\,\text{O}_2 + 56.4\,\text{N}_2$$

(b) Dry products

Water is removed, so the dry mole total is $8 + 2.5 + 56.4 = 66.9$ mol.

Total wet products are 75.9 mol, so the water vapour is $9/75.9 = 11.9\%$ of the wet gas.

Answer: $(A/F)_{stoich} = 15.09$, $(A/F)_{actual} = 18.11$ kg air per kg fuel; dry products 11.96% CO$_2$, 3.74% O$_2$, 84.30% N$_2$.