How much air is needed to burn 3 cm ^ 3 of natural gas, which contains 94%

How much air is needed to burn 3 cm ^ 3 of natural gas, which contains 94% CH4.2% C2H6.1% C4H10.0.5% N2.0.5% CO2. Volume fraction of oxygen in air 0.2

Given:
V (natural gas) = 3 cm3
φ (CH4) = 94%
φ (C2H6) = 2%
φ (C4H10) = 1%
φ (N2) = 0.5%
φ (CO2) = 0.5%
φ (O2) = 0.2

To find:
V (air) -?

1) CH4 + 2O2 => CO2 + 2H2O;
2C2H6 + 7O2 => 4CO2 + 6H2O;
2C4H10 + 13O2 => 8CO2 + 10H2O;
2) V (CH4) = φ (CH4) * V (natural gas) / 100% = 94% * 3/100% = 2.82 cm3;
3) V1 (O2) = V (CH4) * 2 = 2.82 * 2 = 5.64 cm3;
4) V (C2H6) = φ (C2H6) * V (natural gas) / 100% = 2% * 3/100% = 0.06 cm3;
5) V2 (O2) = V (C2H6) * 7/2 = 0.06 * 7/2 = 0.21 cm3;
6) V (C4H10) = φ (C4H10) * V (natural gas) / 100% = 1% * 3/100% = 0.03 cm3;
7) V3 (O2) = V (C4H10) * 13/2 = 0.03 * 13/2 = 0.195 cm3;
8) V total. (O2) = V1 (O2) + V2 (O2) + V3 (O2) = 5.64 + 0.21 + 0.195 = 6.045 cm3;
9) V (air) = V (O2) * 100% / φ (O2) = 6.045 / 0.2 = 30.2 cm3.

Answer: The air volume is 30.2 cm3.