How many grams of copper(II) hydroxide will precipitate when 20mL of .65M sodium hydroxide is mixed with 30mL of .25M copper (II) sulfate.?
2017-11-05 6:11 pm
The correct answer is 0.63 but I keep on arriving at 0.73. Please let me know how to arrive at 0.63. Thank you.
回答 (1)
2017-11-05 6:59 pm
Molar mass of Cu(OH)₂ = (63.5 + 16.0×2 + 1.0×2) g/mol = 97.5 g/mol
Initial no. of moles of CuSO₄ = (0.25 mol/L) × (30/1000 L) = 0.0075 mol
Initial no. of moles of NaOH = (0.65 mol/L) × (20/1000 L) = 0.013 mol
CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
Mole ratio CuSO₄ : NaOH : Cu(OH)₂ = 1 : 2 : 1
When 0.013 mol NaOH completely reacts, CuSO₄ needed = (0.013 mol) × (1/2) = 0.0065 mol < 0.0075 mol
Hence, CuSO₄ is in excess, and thus NaOH is the limiting reactant/reagent.
No. of moles of Cu(OH)₂ precipitated = (0.013 mol) × (1/2) = 0.0065 mol
Mass of Cu(OH)₂ precipitated = (0.0065 mol) × (97.5 g/mol) = 0.634 g
Initial no. of moles of CuSO₄ = (0.25 mol/L) × (30/1000 L) = 0.0075 mol
Initial no. of moles of NaOH = (0.65 mol/L) × (20/1000 L) = 0.013 mol
CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
Mole ratio CuSO₄ : NaOH : Cu(OH)₂ = 1 : 2 : 1
When 0.013 mol NaOH completely reacts, CuSO₄ needed = (0.013 mol) × (1/2) = 0.0065 mol < 0.0075 mol
Hence, CuSO₄ is in excess, and thus NaOH is the limiting reactant/reagent.
No. of moles of Cu(OH)₂ precipitated = (0.013 mol) × (1/2) = 0.0065 mol
Mass of Cu(OH)₂ precipitated = (0.0065 mol) × (97.5 g/mol) = 0.634 g
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