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Question

If you dissolve 25.0 g of CuSO4(s) in water, then add 25.0 g of NH3(aq), what is the theoretical yield of Cu(NH3)4SO4(s)?

Answer 1

CuSO₄(s) + 4NH₃(aq) ==> Cu(NH₃)₄SO₄

moles CuSO₄ present = 25.0 g x 1 mol/159.6 g = 0.1566 moles

moles NH₃ present = 25.0 g x 1 mol/17 g = 1.47 moles

Based on the mole ratio of 1 mol CuSO₄ to 4 moles NH₃, the CuSO₄ will run out first and is LIMITING.

Now, we use the limiting reactant to find the theoretical yield of the product.

0.1566 mol CuSO₄ x 1 mol Cu(NH₃)₄SO₄/mol CuSO₄ x 228 g xCu(NH₃)₄SO₄/mol = 35.7 g Cu(NH₃)₄SO₄