Calculate the volume of carbon dioxide that is produced by burning 20 liters of ethane in 40 liters of oxygen.

1. We write down the equation:
V = 20 HP V = 40 HP X l. -?
2С2Н6 + 7О2 = 4СО2 + 6Н2О + Q – combustion of ethane, heat, carbon dioxide and water are released;
2. Calculation for the starting materials:
1 mol of gas at normal level – 22.4 liters;
X mol (C2H6) -20 l. hence, X mol (C2H6) = 1 * 20 / 22.4 = 0.89 mol (the substance is in short supply);
1 mol of gas at normal level – 22.4 liters;
X mol (O2) – 40 liters. hence, X mol (O2) = 1 * 40 / 22.4 = 1.785 mol (substance in excess);
Calculations are made taking into account the deficient substance.
3. Proportion:
0.89 mol (C2H6) –X mol (CO2);
-2 mol – 4 mol from here, X mol (CO2) = 0.89 * 4/2 = 1.78 mol.
4. Find the volume of CO2:
V (CO2) = 1.78 * 22.4 = 39.87 liters.
Answer: carbon monoxide (4) with a volume of 39.87 liters was formed.