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)

1. How many mL of 0.25 M Na2CO3 would be needed to precipitate all of the copper ions in 32.2 mL of 0.69 M CuSO4?

2. How many grams of CuCO3(s) would be formed from the copper ions in 32.2 mL of 0.69 M CuSO4?

Answer 1

1) Number of moles of CuSO4 = molarity * volume of solution in L

Number of moles of CuSO4 = 0.69 * 0.0322 L = 0.0222 mol

From the balanced equation we can say that

1 mole of CuSO4 requires 1 mole of Na2CO3 so

0.0222 mole of CuSO4 will require 0.0222 mole of Na2CO3

Molarity of Na2CO3 = number of moles of Na2CO3 / volume of solution in L

0.25 M = 0.0222 / volume of solution in L

volume of solution in L = 0.0222 / 0.25 = 0.0888 L

1L = 1000 mL

0.0888 L = 88.8 mL

Therefore, the volume of solution = 88.8 mL

2) Number of moles of CuSO4 = molarity * volume of solution in L

Number of moles of CuSO4 = 0.69 * 0.0322 L = 0.0222 mol

From the balanced equation we can say that

1 mole of CuSO4 produces 1 mole of CuCO3 so

0.0222 mole of CuSO4 will produce 0.0222 mole of CuCO3

Number of moles of CuCO3 = mass of CuCO3 / molar mass of CuCO3

0.0222 mole = mass of CuCO3 / 123.55 g/mol

mass of CuCO3 = 0.0222 mol * 123.55 g/mol = 2.74 g

Therefore, the mass of CuCO3 produced would be 2.74 g