Find the volume of air required to burn 56 g of ethylene.

1. Let’s write down the reaction equation:

C2H4 + 3O2 = 2CO2 + 2H2O.

2. Find the amount of ethylene reacted:

n (C2H4) = m (C2H4) / M (C2H4) = 56 g / 28 g / mol = 2 mol.

3. Using the reaction equation, we find the amount and then the volume of reacted oxygen (Vm is the molar volume, constant equal to 22.4 l / mol):

n (O2) = n (C2H4) * 3 = 2 mol * 3 = 6 mol.

V (O2) = n (O2) * Vm = 6 mol * 22.4 L / mol = 134.4 L.

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

V (air) = V (O2) * 100% / ω (O2) = 134.4 l * 100% / 21% = 640 l.

Answer: V (air) = 640 l.