Calculate the volume of oxygen required for the complete combustion of 74 liters of ethylene?

Let’s execute the solution:
1. Let’s write the equation according to the problem statement:
V = 74 l. X l. -?
C2H4 + 2O2 = 2CO2 + 2H2O + Q – combustion of ethylene is accompanied by the release of heat, carbon dioxide and water;
2. Proportions:
1 mol of gas at normal level – 22.4 liters;
X mol (C2H4) -74 L. hence, X mol (C2H4) = 1 * 74 / 22.4 = 3.3 mol.
3.3 mol (C2H4) – X mol (O2);
-1 mol                  – 3 mol from here, X mol (O2) = 3.3 * 3/1 = 9.9 mol.
4. Find the volume of O2:
V (O2) = 9.9 * 22.4 = 221.76 liters.
Answer: for the combustion of ethylene, oxygen with a volume of 221.76 liters is required.