Carbon monoxide with a volume of 6.72 dm3 was mixed with oxygen weighing 4.8 g.

Carbon monoxide with a volume of 6.72 dm3 was mixed with oxygen weighing 4.8 g. The mixture was set on fire. Determine the mass of the reaction product.

Let’s execute the solution:
1. Let’s write down the equation, select the coefficients:
2СО + О2 = 2СО2 – ОВР, carbon monoxide was obtained (4);
2. Calculation of molar masses:
M (CO) = 28 g / mol;
M (O2) = 32 g / mol;
M (CO2) = 44 g / mol.
3. Determine the number of moles of CO, O2:
V (CO) = 6.72 dm3 = 6.72 l;
1 mol of gas at normal level – 22.4 liters;
X mol (CO) – 6.72 liters. hence, X mol (CO) = 1 * 6.72 / 22.4 = 0.3 mol;
Y (O2) = m / M = 4.8 / 32 = 0.15 mol (deficient substance);
Calculations are made for the substance in deficiency.
0.15 mol (O2) – X mol (CO2);
-1 mol -2 mol from here, Hops (CO2) = 1.15 * 2/1 = 0.3 mol.
4. Find the mass of carbon monoxide (4) by the formula:
m (CO2) = Y * M = 0.3 * 44 = 13.2 g.
Answer: the mass of the reaction product of carbon dioxide is 13.2 g.