past paper question?
2016-07-05 3:25 pm
An 801 mg sample containing sulfate was dissolved in water and then treated with an excess of barium chloride. The precipitate was washed and dried at 800°C to remove all traces of water. The precipitate was found to weigh 377 mg. What was the percentage sulfate in the sample?
回答 (4)
2016-07-05 4:07 pm
Molar mass of BaSO₄ (barium sulfate) = (137.3 + 32.1 + 16.0×4) g/mol = 233.4 g/mol
Molar mass of SO₄²⁻ (sulfate) = (32.1 + 16.0×4) g/mol = 96.1 g/mol
Consider 1 mole of BaSO₄ :
233.4 g of BaSO₄ contains 96.1 g of SO₄²⁻.
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Mass of BaSO₄ precipitate formed = 377 mg
Mass of SO₄²⁻ = (377 mg) × (96.1/233.4) = 155 mg
Mass % of SO₄²⁻ in the sample = (155/801) × 100% = 19.4%
Molar mass of SO₄²⁻ (sulfate) = (32.1 + 16.0×4) g/mol = 96.1 g/mol
Consider 1 mole of BaSO₄ :
233.4 g of BaSO₄ contains 96.1 g of SO₄²⁻.
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Mass of BaSO₄ precipitate formed = 377 mg
Mass of SO₄²⁻ = (377 mg) × (96.1/233.4) = 155 mg
Mass % of SO₄²⁻ in the sample = (155/801) × 100% = 19.4%
2016-07-05 4:02 pm
the % of SO4 in BaSO4 is 96.06/233.40 x 100% = 41.16%
the mass sulfate in the precipitate is 377 x 0.4116 = 155.3 mg
assuming complete precipitation the % sulfate in the sample is then 155.3/801 x 100% = 19.4%
the mass sulfate in the precipitate is 377 x 0.4116 = 155.3 mg
assuming complete precipitation the % sulfate in the sample is then 155.3/801 x 100% = 19.4%
2016-07-05 3:26 pm
mnj
2016-07-05 3:28 pm
~47.07%
收錄日期: 2021-04-18 15:13:52
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