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

In accordance with the condition of the problem, we compose the reaction equation:
С3Н4 + 4О2 = 3СО2 + 2Н2О – propadiene combustion reaction, carbon dioxide and water are released;
M (C3H4) = 40 g / mol; M (O2) = 32 g / mol;
Let’s determine the amount of moles of propadiene, oxygen:
1 mol of gas at n. y – 22.4 l;
X mol (C3H4) -30 L. hence, X mol (C3H4) = 1 * 30 / 22.4 = 1.339 mol;
1.339 mol (C3H4) – X mol (O2);
-1 mol -4 mol from here, X mol (O2) = 1.339 * 4/1 = 5.35 mol;
We find the volume of oxygen:
V (O2) = 5.35 * 22.4 = 119.97 l.
Answer: the reaction requires oxygen with a volume of 119.97 liters.