# Calculate the volume of O2 that is required for the combustion of 8.8 g of propane.

October 2, 2021 | education

| To solve the problem, we compose the equation:

m = 8.8 g. X l. -?

1. С3Н8 + 5О2 = 3СО2 + 4Н2О + Q – propane combustion, carbon monoxide (4), water is formed;

2. Calculations by formulas:

M (C3H8) = 44 g / mol;

M (O2) = 32 g / mol;

Y (C3H8) = m / M = 8.8 / 44 = 0.2 mol.

4. Proportion:

0.2 mol (C3H8) – X mol (O2);

– 1 mol – 5 mol from here, X mol (O2) = 0.2 * 5/1 = 1 mol.

4. Find the volume of O2:

V (O2) = 1 * 22.4 = 22.4 liters.

Answer: to carry out the process, oxygen with a volume of 22.4 liters is required.

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