How much oxygen is required to burn 3.2 g of methane?

The oxidation reaction of methane with oxygen occurs in accordance with the following chemical reaction equation:

CH4 + 2O2 = CO2 + 2H2O;

Combustion of 1 mol of methane requires 2 mol of oxygen.

Let’s find the required amount of methane.

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

N CH4 = 3.2 / 16 = 0.2 mol;

To oxidize such an amount of methane, 0.2 x 2 = 0.4 mol of oxygen is required.

Let’s find its volume.

One mole of gas under normal conditions fills a volume of 22.4 liters.

The oxygen volume will be equal to:

V O2 = 0.4 x 22.4 = 8.96 liters;