What mass of oxygen and air is required for the complete combustion of 48 g of methane CH4?

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

2CH4 + 3O2 = 2CO2 + 2H2O;

For combustion of 2 moles of ethane, 3 moles of oxygen are required.

Let’s define the molar mass of methane:

M CH4 = 12 + 4 x 1 = 16 grams / mol;

Let’s find the amount of substance in 48 grams of methane:

N CH4 = 48/16 = 3 mol;

For a reaction with 3 moles of oxygen, 3 x 3/2 = 4.5 moles of oxygen will be required.

Let’s find the mass of this amount of oxygen:

m O2 = 4.5 x 16 x 2 = 144 grams;

The mass fraction of oxygen in the air is 23.15%.

Let’s find the required mass of air:

m air = 144 / 0.2315 = 622 grams;