Question

There are two possible methods for recovering copper from a solution of copper (II) sulfate, one uses precipitation and the other uses redox.

(a) Copper(II) ions can be precipitated as copper(II) carbonate.

CuSO₄(aq) + Na₂CO₃(aq) → CuCO₃(s) + Na₂SO₄(aq)

How many mL of 0.144 M Na₂CO₃ would be needed to precipitate all of the copper ions in 25.5 mL of 0.165 M CuSO₄? How many grams of CuCO₃ could be recovered? (Include units in your answer. More information.)

Na2CO3	_____
CuCO3	_____

(b) Copper(II) ions can be displaced by aluminum metal.

3 CuSO₄(aq) + 2 Al(s) → Al₂(SO₄)₃(aq) + 3 Cu(s)

How many grams of Al would be needed to displace all of the copper ions in 25.5 mL of 0.165 M CuSO₄? How many grams of Cu could be recovered? (Include units in your answer. More information.)

Al	______
Cu	______

Please be very specific with the answers!! I only have one submission left and im very lost!

Answer 1

a) The balanced equation is

CuSO₄ (aq) + Na₂CO₃ (aq) -------> CuCO₃ (aq) + Na₂SO₄ (aq)

Moles of CuSO₄ present in solution = (volume of CuSO₄ in L)*(concentration of CuSO₄ in moles/L) = (25.5 mL)*(1 L/1000 mL)*(0.165 mole/L) = 4.2075*10^-3 mole.

As per the balanced equation, the molar ratio of reaction between CuSO₄ and Na₂CO₃ is 1:1.

Therefore, moles of Na₂CO₃ required for precipitation = 4.2075*10^-3 mole.

Volume of Na₂CO₃ required for precipitation = moles of Na₂CO₃/concentration of Na₂CO₃ in mole/L = 4.2075*10^-3 mole/(0.144 mole/L) = 0.02921 L = (0.029218 L)*(1000 mL/1 L) = 29.218 mL ≈ 29.2 mL (ans).

Again, refer to the molar equation. The molar ratio of CuSO₄ and CuCO₃ is 1:1.

Therefore, moles of CuCO₃ produced = 4.2075*10^-3 mole.

Molar mass of CuCO₃ = 123.555 g/mol.

Therefore, mass of CuCO₃ recovered = (4.2075*10^-3 mole)*(123.555 g/1 mole) = 0.5198 g ≈ 0.52 g

Ans: Na₂CO₃ = 29.2 mL; CuCO₃ = 0.52 g

b) The balanced equation is

3 CuSO₄ (aq) + 2 Al (s) -------> 3 Cu (s) + Al₂(SO₄)₃ (aq)

As before, moles of CuSO₄ present = 4.2075*10^-3.

The molar ration of CuSO₄ and Al is 3:2.

Therefore, moles of Al required for the displacement = (4.2075*10^-3 mole CuSO₄)*(2 mole Al/3 mole CuSO₄) = 2.805*10^-3 mole.

Molar mass of Al = 26.981 g/mol.

Therefore, mass of Al required = (2.805*10^-3 mole)*(26.981 g/1 mole) = 0.0757 g ≈ 0.076 g.

Again, refer to the molar equation; the ratio of CuSO₄ and Cu is 3:3 or 1:1. Therefore, the moles of Cu displaced = 4.2075*10^-3.

Molar mass of Cu is 63.546 g/mol.

Therefore, mass of Cu displaced = (4.2075*10^-3 mole)*(63.546 g/1 mole) = 0.2673 g ≈ 0.27 g.

Ans: Al = 0.076 g; Cu = 0.27 g