A sample of 3.80g of impure calcium carbonate was allowed to react with excess dilute hydrochloride acid .?
2016-01-26 10:31 am
It was found that 1.43g of carbon dioxide were given offin the reaction.calculate the percentage by mass of calcium carbonate in the sample
回答 (1)
2016-01-28 8:38 am
Molar mass of CaCO₃ = 40 + 12 + 16×3 = 100 g/mol
Molar mass of CO₂ = 12 + 16×2 = 44 g/mol
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Molar ratio CaCO₃ : CO₂ = 1 : 1
No. of moles of CO₂ = 1.43/44 = 0.0325 mol
No. of moles of CaCO₃ = 0.0325
Mass of CaCO₃ = 0.0325 × 100 = 3.25 g
% by mass of CaCO₃ in the sample = (3.25/3.80) × 100% = 85.5%
Molar mass of CO₂ = 12 + 16×2 = 44 g/mol
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Molar ratio CaCO₃ : CO₂ = 1 : 1
No. of moles of CO₂ = 1.43/44 = 0.0325 mol
No. of moles of CaCO₃ = 0.0325
Mass of CaCO₃ = 0.0325 × 100 = 3.25 g
% by mass of CaCO₃ in the sample = (3.25/3.80) × 100% = 85.5%
收錄日期: 2021-04-20 16:15:33
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