Calculate the volume of carbon dioxide that is emitted as a result of the combustion of 20 liters of ethane.

To solve the problem, we write down data on the condition of the problem:

2С2Н6 + 7О2 = 4СО2 + 6Н2О + Q – ethane combustion, carbon dioxide is released;
Proportions:
1 mole of gas at normal level – 22.4 liters;

X mol (C2H6) – 20 liters from here, X mol (C2H6) = 1 * 20 / 22.4 = 0.9 mol;

0.9 mol (C2H6) – X mol (CO2);

– 2 mol – 4 mol from here, X mol (CO2) = 0.9 * 4/2 = 1.8 mol.

Find the volume of the product:
V (CO2) = 1.8 * 22.4 = 40.32 L

Answer: received carbon monoxide with a volume of 40.32 g