How many liters of C3h8 burns in 224 liters of oxygen?

To solve, we compose the equation of the process:

С3Н8 + 5О2 = 3СО2 + 4Н2О + Q – propane combustion, heat, carbon dioxide, water is released;
Calculation:
M (C3H8) = 44 g / mol.

3. Proportions:

1 mol of gas at normal level – 22.4 liters.

X mol (O2) – 224 liters. hence, X mol (O2) = 1 * 224 / 22.4 = 10 mol.

X mol (C3H8) – 10 mol (O2);

-1 mol -5 mol from here, X mol (C3H8) = 1 * 10/5 = 2 mol.

We find the volume of the original substance:
V (C3H8) = 2 * 22.4 = 44.8 liters.

Answer: you need propane with a volume of 44.8 liters.