Calculate the volume of carbon dioxide (CO2) that is formed by the interaction of 200 g of calcium carbonate (CaCO3) containing 20% impurities with hydrochloric acid (HCl).
1. Let’s compose the equality of chemical interaction:
CaCO3 + 2HCl = CaCl2 + CO2 + H2O.
2. Find the amount of calcium carbonate:
ω (CaCO3) = 100% – ω (impurities) = 100% – 20% = 80%.
m (CaCO3) = m (mixtures) * ω (CaCO3) / 100% = 200 g * 80% / 100% = 160 g.
n (CaCO3) = m (CaCO3) / M (CaCO3) = 160 g / 100 g / mol = 1.6 mol.
3. Using the reaction equation, we find the chemical amount of carbon monoxide and, as a result, its volume (Vm – molar volume, constant equal to 22.4 l / mol):
n (CO2) = n (CaCO3) = 1.6 mol.
V (CO2) = n (CO2) * Vm = 1.6 mol * 22.4 l / mol = 35.84 l.
Answer: V (CO2) = 35.84 liters.
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