Question

The reaction of 5.0 g of copper sulfate with excess sodium hydroxide produced 2.6 g of copper hydroxide. What percent yield of copper hydroxide is obtained?

Answer 1

Write the balanced chemical equation for the reaction as

CuSO₄ (aq) + 2 NaOH (aq) -------> Cu(OH)₂ (s) + Na₂SO₄ (aq)

As per the stoichiometric equation,

1 mole CuSO₄ = 1 mole Cu(OH)₂

Molar mass of CuSO₄ = (1*63.546 + 1*32.065 + 4*15.9994) g/mol = 159.6086 g/mol.

Molar mass of Cu(OH)₂ = (1*63.546 + 2*1.008 + 2*15.9994) g/mol = 97.5608 g/mol.

Mole(s) of CuSO₄ corresponding to 5.0 g CuSO₄ = (5.0 g)/(159.6086 g/mol) = 0.031327 mole.

Theoretical mole(s) of Cu(OH)₂ expected = mole(s) of CuSO₄ taken = 0.031327 mole.

Theoretical mass of Cu(OH)₂ expected = (0.031327 mole)*(97.5608 g/mol) = 3.056287 g.

The actual mass of Cu(OH)₂ obtained is 2.6 g; therefore, the percent yield is (2.6 g)/(3.056287 g)*100 = 85.070544% ≈ 85.07% (ans).