How many kilograms of carbon monoxide (IV) should be released when burning 500 kg

How many kilograms of carbon monoxide (IV) should be released when burning 500 kg of limestone containing 92% calcium carbonate?

Given:
m (limestone) = 500 kg = 500000 g
ω (CaCO3) = 92%

To find:
m (CO2) -?

Decision:
1) CaCO3 => CaO + CO2 ↑;
2) m (CaCO3) = ω (CaCO3) * m (limestone) / 100% = 92% * 500,000 / 100% = 460,000 g;
3) n (CaCO3) = m (CaCO3) / M (CaCO3) = 460000/100 = 4600 mol;
4) n (CO2) = n (CaCO3) = 4600 mol;
5) m (CO2) = n (CO2) * M (CO2) = 4600 * 44 = 202400 g = 202.4 kg.

Answer: The mass of CO2 is 202.4 kg.