How much oxygen is needed to burn 11.2 liters of ethylene?

univerkov | December 8, 2020 | education

Reaction equation:
C2H4 + 3O2 = 2CO2 + 2H2O
Let’s calculate the amount of ethylene substance:
n = V / Vm = 11.2 l / 22.4 l / mol = 0.5 mol
According to the reaction equation:
n (O2) = 3n (C2H4) = 1.5 mol
Let’s calculate the volume of 1.5 mol of oxygen:
V = Vm * n = 22.4 l / mol * 1.5 mol = 33.6 l

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