What mass of CaO is obtained by decomposition of 250 g of CaCO3 containing 20% of impurities, how much CO2 is produced?

univerkov | February 5, 2021 | education

1. Let’s write the reaction equation:

CaCO3 = CaO + CO2.

2. Find the amount of pure calcium carbonate:

ω (CaCO3) = 100% – ω (impurities) = 100% – 20% = 80%.

m (CaCO3) = m (solution) * ω (CaCO3) / 100% = 250 g * 80% / 100% = 200 g.

n (CaCO3) = m (CaCO3) / M (CaCO3) = 200 g / 100 g / mol = 2 mol.

3. According to the reaction equation, we find the amount of substances, and then the mass of calcium oxide and the volume of gas (Vm is the molar volume, constant equal to 22.4 l / mol):

n (CaO) = n (CaCO3) = 2 mol.

m (CaO) = n (CaO) * M (CaO) = 2 mol * 56 g / mol = 112 g.

n (CO2) = n (CaCO3) = 2 mol.

V (CO2) = n (CO2) * Vm = 2 mol * 22.4 l / mol = 44.8 l.

Answer: m (CaO) = 112 g; V (CO2) = 44.8 liters.

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