Dissolving 5 g of a mixture of calcium oxide and calcium carbonate in an acid produced 140 ml of gas. Determine the mass fraction of calcium oxide in the initial mixture.
m (mixture) = 5 g
V (gas) = 140 ml = 0.14 l
ω (CaO) -?
1) CaO + 2HCl => CaCl2 + H2O;
CaCO3 + 2HCl => CaCl2 + CO2 ↑ + H2O;
2) n (CO2) = V (CO2) / Vm = 0.14 / 22.4 = 0.00625 mol;
3) n (CaCO3) = n (CO2) = 0.00625 mol;
4) M (CaCO3) = Mr (CaCO3) = Ar (Ca) * N (Ca) + Ar (C) * N (C) + Ar (O) * N (O) = 40 * 1 + 12 * 1 + 16 * 3 = 100 g / mol;
5) m (CaCO3) = n (CaCO3) * M (CaCO3) = 0.00625 * 100 = 0.625 g;
6) m (CaO) = m (mixture) – m (CaCO3) = 5 – 0.625 = 4.375 g;
7) ω (CaO) = m (CaO) * 100% / m (mixture) = 4.375 * 100% / 5 = 87.5%.
Answer: The mass fraction of CaO is 87.5%.
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