Find the volume of carbon dioxide that is released during the combustion of 110m3 of ethylene (C2H2).

In accordance with the condition of the problem, we compose the equation:
2С2Н2 + 5О2 = 4СО2 + 2Н2О – reaction of combustion of ethine, carbon dioxide was obtained;
Let’s convert the volume of acetylene to liters:
1 l. – 0.001 m3;
X – 110 m3 from here X l (C2H2) = 1 * 110 / 0.001 = 110,000 l.
Determine the number of moles of acetylene:
1 mol of gas at normal level – 22.4 liters.
X mol (C2H2) – 110,000 liters. hence, X mol (C2H2) = 1 * 110,000 / 22.4 = 4910.7 mol.
Let’s make the proportion:
4910.7 mol (C2H2) – X mol (CO2);
– 2 mol – 4 mol from here, X mol (CO2) = 4 * 4910.7 / 2 = 9821 mol.
We find the volume of carbon dioxide:
V (CO2) = 9821 * 22.4 = 219990 HP
Answer: The volume of carbon dioxide is 219990 liters.