Question

Calculate the theoretical yield of [Cu(NH₃)₄]SO₄ H₂O from 2.00 g of CuSO₄

CuSO₄ --> Cu²⁺ + SO_4^2-

Cu²⁺ + 4NH₃ --> [Cu(NH₃)₄]²⁺

[Cu(NH₃)₄]²⁺ + SO₄^2- + H₂O --> [Cu(NH₃)₄]SO₄ H₂O

Answer 1

CuSO₄ --> Cu²⁺ + SO₄^2-

Cu²⁺ + 4NH₃ --> [Cu(NH₃)₄]²⁺

[Cu(NH₃)₄]²⁺ + SO₄^2- + H₂O --> [Cu(NH₃)₄]SO₄ H₂O

Amount of CuSO₄ used = 2.00 g

Molecular mass of CuSO₄ = 159.609 g/mole

So number of moles of CuSO₄ used = 2.00 / 159.6 = 0.01253 moles

The balanced chemical equation shows that 1 mole of CuSO₄ gives 1 mole of [Cu(NH₃)₄]SO₄.H₂O

So number of moles of [Cu(NH₃)₄]SO₄.H₂O formed = 0.01253 mole

Molecular mass of [Cu(NH₃)₄]SO₄.H₂O = 245.6 g/mole

The theoritical mass of [Cu(NH₃)₄]SO₄.H₂O formed = 0.01253 mol x 245.6 g/mol = 3.077 g

The theoritical mass of [Cu(NH₃)₄]SO₄.H₂O formed from 2 g of CuSO₄ = 3.077 g