Determine how much carbon monoxide (IV) is released during the combustion of 3.36 liters of ethylene (C2H4).

Let’s implement the solution:

According to the condition of the problem, we will write the data:
V = 3.36 l Chl -?

2С2Н4 + 6О2 = 4СО2 + 4Н2О + Q – combustion of ethylene, carbon dioxide is released;

Proportions:
1 mol of gas at normal level – 22.4 liters;

X mol (C2H4) – 3.36 L from here, X mol (C2H4) = 1 * 3.36 / 22.4 = 0.15 mol;

0.15 mol (C2H4) – X mol (CO2);

-2 mol – 4 mol from here, X mol (CO2) = 0.15 * 4/2 = -0.3 mol.

Find the volume of the product:
V (CO2) = 0.3 * 22.4 = 6.72 L

Answer: received carbon monoxide (4) with a volume of 6.72 liters