Understanding Stoichiometric Air-Fuel Ratio for Petrol Combustion

The stoichiometric air-fuel ratio is a crucial concept in combustion, particularly for engines. It represents the theoretically ideal ratio of air to fuel required for complete combustion. In complete combustion, all the fuel is burned using all the available oxygen, producing only carbon dioxide ($ \text{CO}_2 $) and water ($ \text{H}_2\text{O} $).

Petrol is a complex mixture of hydrocarbons, but it is often approximated as octane ($ \text{C}_8\text{H}_{18} $) for calculations. Air is primarily a mixture of nitrogen ($ \text{N}_2 $) and oxygen ($ \text{O}_2 $). For combustion purposes, air is considered to be approximately 23.2% oxygen and 76.8% nitrogen by mass.

Calculating Stoichiometric Ratio

The balanced chemical equation for the complete combustion of octane ($ \text{C}_8\text{H}_{18} $) with oxygen is:

$ \text{C}_8\text{H}_{18} + 12.5 \text{O}_2 \rightarrow 8 \text{CO}_2 + 9 \text{H}_2\text{O} $

This equation tells us that one mole of octane requires 12.5 moles of oxygen for complete combustion.

To find the ratio by mass, we use the molar masses:

• Molar mass of C = 12.011 g/mol
• Molar mass of H = 1.008 g/mol
• Molar mass of O = 15.999 g/mol

• Molar mass of $ \text{C}_8\text{H}_{18} = (8 \times 12.011) + (18 \times 1.008) \approx 96.088 + 18.144 = 114.232 $ g/mol
• Molar mass of $ \text{O}_2 = 2 \times 15.999 = 31.998 $ g/mol

Mass of oxygen required for 1 mole of octane = $ 12.5 \times 31.998 \approx 399.975 $ g

Since air is approximately 23.2% oxygen by mass, the mass of air required is:

Mass of air = $ \frac{\text{Mass of } \text{O}_2}{\text{Fraction of } \text{O}_2 \text{ in air}} = \frac{399.975}{0.232} \approx 1724.03 $ g

The stoichiometric air-fuel ratio by mass is:

Air/Fuel Ratio = $ \frac{\text{Mass of air}}{\text{Mass of fuel}} = \frac{1724.03}{114.232} \approx 15.09 $

This calculated value for octane is approximately 15.09:1. For typical petrol, which is a mixture of hydrocarbons, the stoichiometric air-fuel ratio by mass is commonly accepted to be around 14.7:1. However, in simplified contexts or for specific calculations depending on the exact fuel composition, this value is often rounded or approximated.

Let's look at the given options:

• 5 : 1 (This is a very fuel-rich mixture)
• 10 : 1 (This is also a fuel-rich mixture)
• 12 : 1 (This is still a fuel-rich mixture)
• 15 : 1 (This is close to the theoretically calculated value and the commonly cited value for petrol)

The value 15:1 is the closest option to the actual stoichiometric air-fuel ratio for petrol by mass.

Revision Table: Key Terms

Additional Information: Fuel Mixtures

While the stoichiometric ratio is ideal for complete combustion, engines often operate at slightly different air-fuel ratios depending on the requirement:

• Lean Mixture: An air-fuel ratio higher than stoichiometric (e.g., 16:1 or more). There is excess air. This can improve fuel economy but might reduce power and increase nitrogen oxide ($ \text{NO}_\text{x} $) emissions.
• Rich Mixture: An air-fuel ratio lower than stoichiometric (e.g., 13:1 or less). There is excess fuel. This is often used for maximum power output or to cool engine components, but it leads to higher fuel consumption and increased emissions of unburned hydrocarbons (HC) and carbon monoxide (CO).

Maintaining a ratio close to stoichiometric (around 14.7:1 for petrol) is crucial for the efficient operation of catalytic converters, which are used to reduce harmful emissions from exhaust gases.