Calculate the volume of oxygen required for complete combustion of 5.6 liters of propyne.

Let’s execute the solution:
1. According to the condition, we write: С3Н4 + 4О2 = 3СО2 + 2Н2О + Q – propyne combustion, heat, carbon monoxide, water is released;
2. Calculations by formulas:
M (C3H4) = 40 g / mol.
M (O2) = 32 g / mol.
3. Proportions:
1 mol of gas at normal level – 22.4 liters.
X mol (C3H4) -5.6 l. hence, X mol (C3H4) = 1 * 5.6 / 22.4 = 0.25 mol.
0.25 mol (C3H4) – X mol (CO2);
-1 mol -4 mol from here, X mol (CO2) = 0.25 * 4/1 = 1 mol.
4. Find the volume of O2:
V (O2) = 1 * 22.4 = 22.4 liters.
Answer: for the combustion process, oxygen with a volume of 22.4 liters is required.