How much oxygen is needed to burn 30 liters of propadiene?

Let’s implement the solution:
1. Let us write down the reaction equation for the combustion of a diene hydrocarbon:
Н2С = С = СН2 + 4О2 = 3СО2 + 2Н2О + Q – during the combustion of propadiene carbon dioxide, water, heat are released;
2. Determine the amount of moles of the substance – C3H4 (propadiene), oxygen:
1 mole of gas at normal level – 22.4 liters;
X mol (C3H4) – 30 liters. hence, X mol (C3H4) = 1 * 30 / 22.4 = 1.33 mol;
1.33 mol (C3H4) – X mol (O2);
-1 mol -4 mol from here, X mol (O2) = 1.33 * 4/1 = 5.32 mol.
3. We calculate the volume of О2:
V (O2) = 5.32 * 22.4 = 119.17 liters.
Answer: for the combustion reaction, oxygen with a volume of 119.17 liters is required.