What volume of carbon monoxide (IV) will be released during the decomposition of 750 g

What volume of carbon monoxide (IV) will be released during the decomposition of 750 g of known ca, containing 20% of impurities?

1. Let us write down the equation for the decomposition of calcium carbonate:

CaCO3 = CaO + CO2;

2.calculate the mass of calcium carbonate:

m (CaCO3) = m (limestone) – m (impurities);

m (impurities) = w (impurities) * m (limestone) = 0.2 * 750 = 150 g;

m (CaCO3) = 750 – 150 = 600 g;

3. find the chemical amount of carbonate:

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

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

n (CaCO3) = 600: 100 = 6 mol;

4.determine the amount of released carbon dioxide:

n (CO2) = n (CaCO3) = 6 mol;

5.Calculate the volume of CO2:

V (CO2) = n (CO2) * Vm = 6 * 22.4 = 134.4 liters.

Answer: 134.4 liters.