How much air will go to the combustion of one mole of Methane?

The methane oxidation reaction is described by the following chemical reaction equation.
CH4 + 2O2 = CO2 + 2H2O;
According to the coefficients of this equation, 2 oxygen molecules are required to oxidize 1 methane molecule. This synthesizes 1 molecule of carbon dioxide.
The molar volume of the gas is standard and under normal conditions it fills a volume of 22.4 liters.
Since the molar volume of the gas is standard, twice the amount of oxygen will require twice the volume of oxygen.
The amount of oxygen will be.
N O2 = 1 x 2 = 2 mol; Let’s calculate the gas volume. To do this, multiply the amount of substance by the volume of 1 mole of gas.
Its volume will be: V O2 = 2 x 22.4 = 44.8 liters; With the condition that the volume fraction of oxygen in the air is 21%, the required volume of air will be 44.8 / 0.21 = 213.3 liters;