How much air is required to burn a mixture of methane with a volume of 5 liters

How much air is required to burn a mixture of methane with a volume of 5 liters, with ethane with a volume of 2 liters? The volume fraction of oxygen in the air is 21%

Given:
V (CH4) = 5 l
V (C2H6) = 2 L
φ (O2) = 21%

To find:
V (air) -?

1) Write the reaction equations:
CH4 + 2O2 => CO2 + 2H2O;
2C2H6 + 7O2 => 4CO2 + 6H2O;
2) Calculate the amount of O2 spent on CH4 combustion:
V1 (O2) = V (CH4) * 2 = 5 * 2 = 10 L;
3) Calculate the volume of O2 spent on combustion of C2H6:
V2 (O2) = V (C2H6) * 7/2 = 2 * 7/2 = 7 L;
4) Calculate the total O2 volume:
V total (O2) = V1 (O2) + V2 (O2) = 10 + 7 = 17 l;
5) Calculate the air volume:
V (air) = V (O2) * 100% / φ (O2) = 17 * 100% / 21% = 81 liters.

Answer: The air volume is 81 liters.