How many grams of CuCO3 could be recovered?
(a) Copper(II) ions can be precipitated as copper(II) carbonate.
CuSO4(aq) + Na2CO3(aq) → CuCO3(s) + Na2SO4(aq
How many mL of 0.126 M Na2CO3 would be needed to precipitate all of the copper ions in 24.4 mL of 0.163 M CuSO4? H?
2016-11-01 12:27 pm
回答 (1)
2016-11-01 12:40 pm
(a)
CuSO₄(aq) + Na₂CO₃(aq) → CuCO₃(s) + Na₂SO₄(aq)
OR: Mole ratio CuSO₄ : Na₂CO₃ = 1 : 1
No. of milli-moles of CuSO₄ reacted = (0.163 mmol/mL) × (24.4 mL) = 3.98 mmol
No. of milli-moles of Na₂CO₃ needed = 3.98 mmol
Volume of Na₂CO₃ needed = (3.98 mmol) / (0.126 mmol/mL) = 31.6 mL
Refer to the above equation. Mole ratio CuSO₄ : CuCO₃ = 1 : 1
No. of moles of CuSO₄ reacted = 3.98 mmol = 0.0398 mol
No. of moles of CuCO₃ recovered = 0.00398 mol
Molar mass of CuCO₃ = (63.5 + 12.0 + 16.0×3) g/mol = 123.5 g/mol
Molar mass of CuCO₃ recovered = (0.00398 mol) × (123.5 g/mol) = 0.492 g
CuSO₄(aq) + Na₂CO₃(aq) → CuCO₃(s) + Na₂SO₄(aq)
OR: Mole ratio CuSO₄ : Na₂CO₃ = 1 : 1
No. of milli-moles of CuSO₄ reacted = (0.163 mmol/mL) × (24.4 mL) = 3.98 mmol
No. of milli-moles of Na₂CO₃ needed = 3.98 mmol
Volume of Na₂CO₃ needed = (3.98 mmol) / (0.126 mmol/mL) = 31.6 mL
Refer to the above equation. Mole ratio CuSO₄ : CuCO₃ = 1 : 1
No. of moles of CuSO₄ reacted = 3.98 mmol = 0.0398 mol
No. of moles of CuCO₃ recovered = 0.00398 mol
Molar mass of CuCO₃ = (63.5 + 12.0 + 16.0×3) g/mol = 123.5 g/mol
Molar mass of CuCO₃ recovered = (0.00398 mol) × (123.5 g/mol) = 0.492 g
收錄日期: 2021-04-18 15:44:38
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