How much oxygen is required to burn 20 liters of ethine?

To solve the problem, let’s compose the equation:
2С2Н2 + 5О2 = 4СО2 + 2Н2О – reaction of combustion of ethine, carbon dioxide and water are released;
Let’s calculate the molecular weights:
M (C2H2) = 26 g / mol;
M (O2) = 32 g / mol.
Let us determine the amount of moles of ethine, if its volume is known:
1 mol of gas at normal level – 22.4 liters.
X mol (C2H2) -20 L from here, X mol (C2H2) = 1 * 20 / 22.4 = 0.89 mol.
We make the proportion according to the equation:
0.89 mol (C2H2) – X mol (O2);
-2 mol – 5 mol from here, X mol (O2) = 0.89 * 5/2 = 2.225 mol.
Find the volume of oxygen by applying Avogadro’s law:
V (O2) = 2.225 * 22.4 = 49.84 liters.
Answer: for the combustion reaction, oxygen with a volume of 49.84 liters is required.