How much carbon dioxide is released during the decomposition of 250g. limestone containing 20% impurities?

Given:
m (limestone) = 250 g
ω approx. = 20%

To find:
V (CO2) -?

Decision:
1) CaCO3 = (tOC) => CaO + CO2 ↑;
2) M (CaCO3) = Mr (CaCO3) = Ar (Ca) + Ar (C) + Ar (O) * 3 = 40 + 12 + 16 * 3 = 100 g / mol;
3) ω (CaCO3) = 100% – ω approx. = 100% – 20% = 80%;
4) m (CaCO3) = ω (CaCO3) * m (limestone) / 100% = 80% * 250/100% = 200 g;
5) n (CaCO3) = m (CaCO3) / M (CaCO3) = 200/100 = 2 mol;
6) n (CO2) = n (CaCO3) = 2 mol;
7) V (CO2) = n (CO2) * Vm = 2 * 22.4 = 44.8 liters.

Answer: The CO2 volume is 44.8 liters.