How much oxygen and air is required to burn 20 liters of methane?

The methane combustion reaction is described by the following chemical reaction equation:

CH4 + 2O2 = CO2 + 2H2O;

Combustion of 1 mole of methane requires 2 moles of oxygen. This produces 1 mole of carbon dioxide and 2 moles of water.

Under normal conditions, one mole of ideal gas occupies a fixed volume of 22.4 liters.

Therefore, for the combustion of 20 liters of methane, 20 x 2 = 40 liters of oxygen is required.

The oxygen content in the air is 20.95%.

In order to find the required volume of air, we divide the required volume of oxygen by its volume fraction.

The required air volume will be:

V air = 40 / 0.2095 = 191 liters;