What mass of chalk containing 22% impurities was fired if 11.2 liters of gas were released?

In accordance with the condition, we compose the equation for the calcination of calcium carbonate:

CaCO3 = CaO + CO2 – decomposition, carbon dioxide is released;
Calculations:
M (CaCO3) = 100 g / mol;

M (CO2) = 44 g / mol.

3. Proportion:

1 mol of gas at normal level – 22.4 liters;

X mol (CO2) – -11.2 L hence, X mol (CO2) = 1 * 11.2 / 22.4 = 0.5 mol.

Y (CaCO3) = 0.5 mol since the amount of substances is 1 mol.

4. Determine the mass of the original substance:

m (CaCO3) = Y * M = 0.5 * 100 = 50 g (theoretical weight);

m (CaCO3) = 50 * (1 – 0.22) = 39 g (mass without impurities).

Answer: the mass of chalk is 39 g.