Stoichiometric air-fuel ratio refers to the ideal ratio of air to fuel that is required for complete combustion. It is the point at which there is just enough air to burn all the fuel without any excess or deficiency. In the case of methane (CH4) combustion in air, the stoichiometric air-fuel ratio can be calculated using the balanced chemical equation for the combustion reaction.

The balanced chemical equation for the combustion of methane can be written as:

CH4 + 2(O2 + 3.76N2) → CO2 + 2H2O + 7.52N2

From the equation, we can see that one molecule of methane reacts with two molecules of oxygen and 3.76 molecules of nitrogen. This is because air is composed of approximately 21% oxygen (O2) and 79% nitrogen (N2) by volume.

To determine the stoichiometric air-fuel ratio, we need to compare the volumes of air and fuel consumed in the reaction. Since methane is a gas, its volume can easily be measured and compared to the volume of air.

The stoichiometric air-fuel ratio is calculated as follows:

Stoichiometric air-fuel ratio = Volume of air ÷ Volume of fuel

In this case, the volume of air is 1 + (2 × 3.76) = 1 + 7.52 = 8.52

Therefore, the stoichiometric air-fuel ratio is 8.52:1.

However, the given options do not match this calculated value. It is possible that the options provided in the question are approximations or rounded values. Among the given options, option C (9.52:1) is the closest to the calculated stoichiometric air-fuel ratio of 8.52:1.

It is important to note that the stoichiometric air-fuel ratio may vary depending on the specific composition of the fuel and the oxidizer used.