# What is the volume of carbon dioxide released during the decomposition of 400 g of limestone

What is the volume of carbon dioxide released during the decomposition of 400 g of limestone, the mass fraction of impurities in which is 20%?

1. Let’s write down the reaction equation:

CaCO3 = CaO + CO2.

2. Find the amount of calcium carbonate:

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

m (CaCO3) = (m (mixture) * ω (CaCO3)) / 100% = 400 g * 80% / 100% = 320 g.

n (CaCO3) = m (CaCO3) / M (CaCO3) = 320 g / 100 g / mol = 3.2 mol.

3. According to the reaction equation, we find the amount, and then the volume of carbon dioxide (Vm – molar volume, constant equal to 22.4 l / mol):

n (CO2) = n (CaCO3) = 3.2 mol.

V (CO2) = n (CO2) * Vm = 3.2 mol * 22.4 l / mol = 71.68 l.

Answer: V (CO2) = 71.68 liters.

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