Find the volume of air required to burn 80 liters of ethane.

1. Let’s write down the reaction equation:

2C2H6 + 7O2 = 4CO2 + 6H2O.

2. Find the amount of ethane (Vm – molar volume, constant equal to 22.4 l / mol):

n (C2H6) = V (C2H6) / Vm = 80 L / 22.4 L / mol = 3.57 mol.

3. According to the reaction equation, we find the amount, and then the volume, oxygen:

n (O2) = n (C2H6) * 7/2 = 3.57 mol * 7/2 = 12.5 mol.

V (O2) = n (O2) * Vm = 12.5 l * 22.4 l / mol = 280 l.

4. Find the volume of air required for the reaction:

V (air) = V (O2) * 100% / ω (O2) = 280 l * 100% / 21% = 1333.33 l.

Answer: V (air) = 1333.33 liters.