Chemistry question?
2019-03-07 12:06 am
A 5.455-g sample of impure CaCl2 is dissolved and treated with excess potassium carbonate solution. The dried CaCO3 (calcium carbonate) precipitate weighs 4.010-g. Calculate the percent by mass of CaCl2 in the original mixture.
回答 (1)
2019-03-07 12:23 am
✔ 最佳答案
Molar mass of CaCl₂ = (40.08 + 35.45×2) g/mol = 110.98 g/molMolar mass of CaCO₃ = (40.08 + 12.01 + 16.00×3) g/mol = 100.09 g/mol
CaCl₂(aq) + K₂CO₃(aq) → CaCO₃(s) + 2KCl(aq)
Mole ratio CaCl₂ : CaCO₃ = 1 : 1
Moles of CaCO₃ produced = (4.010 g) / (100.09 g/mol) = 0.04006 mol
Moles of CaCl₂ reacted = 0.04006 mol
Mass of CaCl₂ in the sample = (0.04006 mol) × (110.98 g/mol) = 4.446 g
% by mass of CaCl₂ in the sample = (4.446/5.455) × 100% = 81.50%
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