---
title: "Combustion of octane with excess air"
subject: "Internal Combustion Engines"
chapter: 3
key: "semester_6/internal_combustion_engines:octane-combustion-air-fuel-numerical"
source: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/octane-combustion-air-fuel-numerical
site: BCT Engineering
---

# Combustion of octane with excess air

Internal Combustion Engines · Chapter 3: Engine fuels

- Asked in: year not given, 6 marks
- Video: [Mechanical Engineering Thermodynamics - Lec 32, pt 1 of 3:  Combustion - Excess Air](https://www.youtube.com/watch?v=iKJs5Npr-2g) — Ron Hugo
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/octane-combustion-air-fuel-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/internal_combustion_engines/chapter-3

## Question

Octane (C$_8$H$_{18}$) is burned with 20% excess air. Write the balanced combustion equation and calculate (a) the stoichiometric and actual air-fuel ratio by mass, and (b) the volumetric composition of the dry products. Take air as 21% O$_2$ and 79% N$_2$ by volume (3.76 mol N$_2$ per mol O$_2$), molecular masses: C$_8$H$_{18}$ 114.23, air 28.97.

## Short answer

Stoichiometric A/F = 15.09, actual A/F with 20% excess air = 18.11. Dry exhaust composition: 11.96% CO2, 3.74% O2, 84.30% N2 by volume.

## Answer

#### 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.

| Gas | Moles | % by volume (dry) |
|---|---|---|
| CO$_2$ | 8.0 | 11.96 |
| O$_2$ | 2.5 | 3.74 |
| N$_2$ | 56.4 | 84.30 |
| Total | 66.9 | 100 |

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$_2$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How many moles of O2 are needed to burn one mole of C8H18 stoichiometrically?
   A. 12.5
   B. 8
   C. 9
   D. 25
   Answer: A. 8 mol for carbon (to CO2) plus 4.5 mol for 9 H2O gives 12.5. (Key point: Oxygen requirement)

2. What is the stoichiometric air-fuel ratio by mass?
   A. 18.11
   B. 15.09
   C. 12.5
   D. 59.6
   Answer: B. 12.5 x 4.76 x 28.97 / 114.23 = 15.09. (Key point: Stoichiometric A/F)

3. With 20% excess air, what is the actual A/F?
   A. 0.8 x stoichiometric
   B. stoichiometric + 20
   C. 1.2 x stoichiometric = 18.11
   D. stoichiometric / 1.2
   Answer: C. Air supplied is 120% of theoretical, so 1.2 x 15.09 = 18.11. (Key point: Actual A/F with excess air)

4. Which species make up the dry products, and why is the O2 present?
   A. CO2, H2O and N2; no O2
   B. CO2 and N2 only
   C. CO, N2 and H2
   D. CO2, N2 and O2; O2 is the excess air oxygen of 2.5 mol
   Answer: D. Water is removed from dry analysis, and the excess air leaves 15 - 12.5 = 2.5 mol of O2. (Key point: Dry products)

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