What volume of air is required for the complete combustion of 20 liters of a mixture in which the mass

What volume of air is required for the complete combustion of 20 liters of a mixture in which the mass fraction of propane is 10%, butane is 90%.

Given:
V (mixture) = 20 l
φ (C3H8) = 10%
φ (C4H10) = 90%

To find:
V (air) -?

1) C3H8 + 5O2 => 3CO2 + 4H2O;
2C4H10 + 13O2 => 8CO2 + 10H2O;
2) V (C3H8) = φ (C3H8) * V (mixture) / 100% = 10% * 20/100% = 2 l;
3) V1 (O2) = V (C3H8) * 5 = 2 * 5 = 10 L;
4) V (C4H10) = φ (C4H10) * V (mixtures) / 100% = 90% * 20/100% = 18 l;
5) V2 (O2) = V (C4H10) * 13/2 = 18 * 13/2 = 117 l;
6) V total. (O2) = V1 (O2) + V2 (O2) = 10 + 117 = 127 l;
7) V (air) = V (O2) * 100% / φ (O2) = 127 * 100% / 21% = 605 HP.

Answer: The air volume is 605 liters.