How much air is needed to burn methane with the amount of 0.25 mol.

Given:
v (CH4) = 0.25 mol.
interaction of methane (CH4) with oxygen O2;
Find: V (air) -?
Decision:
1) Let’s write the reaction equation:
CH4 + 2O2 = CO2 + 2H2O + Q;
From the reaction – 2 mol of O2 is consumed per 1 mol of CH4;
2) Since v (CH4): v (O2) = 1: 2, then v (O2) = 2v (CH4) = 2 * 0.25 = 0.5 mol;
4) Let us find the volume for oxygen according to Avogadro’s law, while Vm = 22.4 l / mol. V (O2) = 22.4 * 0.5 = 11.2 l.
5) Since there is 21% oxygen in the air, then
V (air) – 100%;
11.2 L – 21%;
V (air) = 11.2 * 100/21 = 53.3 l.