Find the volume of gas that is released during the interaction of 200 g of limestone containing 5% impurities with hydrochloric acid.

1. The reaction proceeds:
CaCO3 + 2HCl = CaCl2 + H2O + CO2 ↑;
2.Calculate the mass of calcium carbonate:
m (CaCO3) = m (limestone) – m (impurities);
m (impurities) = w (impurities) * m (limestone) = 0.05 * 200 = 10 g;
m (CaCO3) = 200-10 = 190 g;
3.Calculate the chemical amounts of calcium carbonate and carbon dioxide:
n (CaCO3) = m (CaCO3): M (CaCO3);
M (CaCO3) = 40 + 12 + 3 * 16 = 100 g / mol;
n (CaCO3) = 190: 100 = 1.9 mol;
n (CO2) = n (CaCO3) = 1.9 mol;
4. find the volume of gas:
V (CO2) = n (CO2) * Vm = 1.9 * 22.4 = 42.56 liters.
Answer: 42.56 liters.