Question

There are two possible methods for recovering copper from a solution of copper (II) sulfate, one uses precipitation and the other uses redox. Copper(II) ions can be precipitated as copper(II) carbonate. CuSO4(aq) + Na2CO3(aq) → CuCO3(s) + Na2SO4(aq) How many mL of 0.363 M Na2CO3 would be needed to precipitate all of the copper ions in 47.0 mL of 0.323 M CuSO4? mL How many grams of CuCO3(s) would be formed from the copper ions in 47.0 mL of 0.323 M CuSO4? g

Answer 1

Given reaction is CuSO₄(aq) + Na₂CO₃(aq) → CuCO₃(s) + Na₂SO₄(aq)

Q1: How many mL of 0.363 M Na₂CO₃ would be needed to precipitate all of the copper ions in 47.0 mL of 0.323 M CuSO₄? in mL

Answer: we can use M1V1 = M2V2 formula here

M1 = 0.363 M; V1 = to be calc; M2 = 0.323 M; V2 = 47.0 mL

V1 = (M2V2) / M1 = (0.323M x 47 mL) / 0.363M = 41.82 mL

Volume of Na₂CO₃ required = 41.8 mL ( 3 sig fig)

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Q2: How many grams of CuCO₃(s) would be formed from the copper ions in 47.0 mL of 0.323 M CuSO₄? in g

Answer: 1 mol CuSO₄ gives 1 mol CuCO₃

moles CuSO₄ = 47.0 mL x 10^-3 L x 0.323 M = v mol

convert mol CuCO₃ into mass CuCO₃ by multiplying mol CuCO₃ with its molar mass

mass of CuCO₃ formed = 0.015181 mol x 123.55 g/mol = 1.87561255 g ~ 1.876 ~ 1.88

grams of CuCO₃(s) would be formed = 1.88 g (3 sig fig)

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