How much oxygen and air is required to burn 2 liters of methane?

1. We compose the reaction equation:

CH4 + 2 O2 = CO2 + 2 H2O

According to the equation:

methane – 1 mol, i.e. 22.4 l

oxygen – 2 mol, i.e. 44.8 l

2. We calculate the volume of oxygen, making up the proportion:

2 l CH4 – xl O2

22.4 L CH4 – 44.8 L O2

Hence, x = 2 * 44.8 / 22.4 = 4 liters.

3. Calculate the volume of air (it contains 21% oxygen):

4 l O2 – 21%

xl of air – 100%

Hence, x = 4 * 100/21 = 19.05 liters.