Calculate what volume of carbon monoxide (IV) can be obtained by thermal decomposition of 200 kg

Calculate what volume of carbon monoxide (IV) can be obtained by thermal decomposition of 200 kg of limestone, the mass fraction of impurities in which is 10%.

Given:
m (limestone) = 200 kg = 200000 g
ω approx. = 10%

To find:
V (CO2) -?

Decision:
1) CaCO3 => CaO + CO2 ↑;
2) ω (CaCO3) = 100% – ω approx. = 100% – 10% = 90%;
3) m (CaCO3) = ω (CaCO3) * m (limestone) / 100% = 90% * 200,000 / 100% = 180,000 g;
4) n (CaCO3) = m (CaCO3) / M (CaCO3) = 180000/100 = 1800 mol;
5) n (CO2) = n (CaCO3) = 1800 mol;
6) V (CO2) = n (CO2) * Vm = 1800 * 22.4 = 40320 l.

Answer: The volume of CO2 is 40320 liters.