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

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.