How many kilograms of carbon monoxide (IV) should be released when burning 500 kg of limestone containing 92% calcium carbonate?
m (limestone) = 500 kg = 500000 g
ω (CaCO3) = 92%
m (CO2) -?
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.
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