Find the volume of air (21% o2 in air) that will be required to burn 224 liters of c3h8 propane.

1. Let’s write down the reaction equation:

C3H8 + 5O2 = 3CO2 + 4H2O.

2. Find the amount of propane that has reacted (Vm is the molar volume, constant equal to 22.4 l / mol):

n (C3H8) = V (C3H8) / Vm = 224 L / 22.4 L / mol = 10 mol.

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

n (O2) = n (C3H8) * 5 = 10 mol * 5 = 50 mol.

V (O2) = n (O2) * Vm = 50 mol * 22.4 L / mol = 1120 L.

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

V (air) = V (O2) * 100% / ω (O2) = 1120 l * 100% / 21% = 5333.33 l.

Answer: V (air) = 5333.33 liters.