15 liters of ethylene burned out, find the volume of carbon dioxide produced.

To solve the problem, it is necessary to compose the reaction equation:
С2Н4 + 3О2 = 2СО2 + 2Н20 – reaction of ethylene combustion, carbon dioxide and water are released;
M (CO2) = 12 + 16 * 2 = 44 g / mol;
Let’s calculate the number of moles of ethylene:
1 mol of gas at n. y occupies a volume of 22.4 liters;
X mol (C2H4) – 15 liters. hence, X mol (C2H4) = 1 * 15 / 22.4 = 0.669 mol;
Let’s make the proportion:
0.669 mol (C2H4) – X mol (CO2);
-1 mol – 2 mol hence, X mol (CO2) = 2 * 0.669 / 1 = 1.338 mol;
Let’s calculate the volume of carbon dioxide:
V (CO2) = 1.338 * 22.4 = 29.97 liters.
Answer: during the reaction, carbon monoxide with a volume of 29.97 liters is released.