A mixture of 20 l of CO and 20 g of O2 burned out. Determine the volume of the generated gas.

Let’s execute the solution:
1. We make the equation:
V = 20 HP m = 20 g. X l. -?
2СО + О2 = 2СО2 – ОВР, carbon monoxide was obtained (4);
2. Proportions:
1 mole of gas at normal level – 22.4 liters;
X mol (CO) – 20 liters. hence, X mol (CO) = 1 * 20 / 22.4 = 0.89 mol (substance in excess);
0.89 mol (CO) – X mol (O2);
-2 mol -1 mol from here, X mol (O2) = 0.89 * 1/2 = 0.446 mol (deficient substance);
Calculations are made for the substance in deficiency.
0.446 mol (O2) – X mol (CO2);
-1 mol – 2 mol from here, X mol (CO2) = 0.446 * 2/1 = 0.89 mol.
3. Find the volume of CO2:
V (CO2) = 0.89 * 22.4 = 19.94 liters.
Answer: carbon monoxide was obtained with a volume of 19.94 liters.