20 g of calcium carbonate was treated with a solution containing 6.3 g of nitric acid

20 g of calcium carbonate was treated with a solution containing 6.3 g of nitric acid. Determine the amount of evolved carbon monoxide (IV).

To solve the problem, we write down the equation of the process:

CaCO3 + 2HNO3 = Ca (NO3) 2 + CO2 + H2O – ion exchange, carbon monoxide is released (4);
Calculations by formulas:
M (CaCO3) = 100 g / mol;

M (CO2) = 44 g / mol;

M (HNO3) = 63 g / mol;

Y (CaCO3) = m / M = 20/100 = 0.2 mol (substance in excess);

Y (HNO3) = m / M = 6.3 / 63 = 0.1 mol (deficient substance).

Calculations are made for the substance in deficiency.

3. Proportion:

0.1 mol (HNO3) – X mol (CO2);

-2 mol -1 mol from here, X mol (CO2) = 0.1 * 1/2 = 0.05 mol.

Find the volume of the oxide:
V (CO2) = 0.05 * 22.4 = 1.12 L

Answer: received carbon monoxide (4) with a volume of 1.12 liters