How much oxygen is required for the complete combustion of 3L propene?

Based on the data of the problem, we write down the equation for the combustion of propene:
2С3Н8 + 9О2 = 6СО2 + 6Н2О – carbon dioxide, water are released;

Calculations by formulas:
M (C3H6) = 42 g / mol;

M (O2) = 32 g / mol.

Proportions:
1 mole of gas at normal level – 22.4 liters;

X mol (C3H6) – 3 liters from here, X mol (C3H6) = 1 * 3 / 22.4 = 0.13 mol.

0.13 mol (C3H6) – X mol (O2);

-2 mol – 9 mol from here, X mol (O2) = 0.13 * 9/2 = 0.585 mol.

We find the volume of O2:
V (O2) = 0.585 * 22.4 = 13.1 L

Answer: you need oxygen in a volume of 13.1 liters