How much carbon dioxide will be released when 5 grams of ethylene are burned?

To solve the problem, let’s compose the reaction equation:
С2Н4 + 3О2 = 2СО2 + 2Н2О – the reaction of ethylene combustion proceeds with the release of carbon dioxide and water;
M (C2H4) = 28 g / mol; M (CO2) = 44 g / mol;
Determine the amount of moles of ethylene by the formula:
Y (C2H4) = m / M = 5/28 = 0.178 mol;
Let’s make the proportion:
0.178 mol (C2H4) – X mol (CO2);
-1 mol -2 mol hence, X mol (CO2) = 0.178 * 2/1 = 0.357 mol;
Let’s calculate the volume of carbon monoxide:
V (CO2) = 0.357 * 22.4 = 7.99 liters.
Answer: 7.99 liters of carbon monoxide was released.