What mass of limestone containing 96% calcium carbonate must be taken for calcination

What mass of limestone containing 96% calcium carbonate must be taken for calcination in order to obtain 672 liters of carbon dioxide.

1. Let us write down the equation of the reaction that occurs during the calcination of calcium carbonate:

CaCO3 = CaO + CO2 ↑;

2.Set the chemical amount of carbon dioxide:

n (CO2) = V (CO2): Vm = 672: 22.4 = 30 mol;

3. Determine the amount of decomposed calcium carbonate:

n (CaCO3) = n (CO2) = 30 mol;

4. find the mass of carbonate:

m (CaCO3) = n (CaCO3) * M (CaCO3);

M (CaCO3) = 40 + 12 + 3 * 16 = 100 g / mol;

m (CaCO3) = 30 * 100 = 3000 g = 3 kg;

5.Calculate the mass of limestone:

m (limestone) = m (CaCO3): w (CaCO3) = 3: 0.96 = 3.125 kg.

Answer: 3.125 kg.