Limiting Reagent

Question

If 10.45 g of aluminum are reacted with 66.55 g of copper (II) sulphate, aluminum sulphate and copper are formed.

a)     Which reactant is in excess?

b)     Calculate the mass of copper formed.

Answer 1

See answer to your previous question about Zn + HCl to see how to find the limiting reagent.

2Al(s) + 3CuSO₄(aq) ==> 3Cu(s) + Al₂(SO₄)₃(aq) ... balanced equation

Limiting reagent:

mols Al = 10.45 g Al x 1 mol Al / 26.98 g = 0.3873 mols Al (÷2->0.194)

mols CuSO₄ = 66.55 g CuSO₄ x 1 mol CuSO₄ / 159.6 g = 0.4170 mols CuSO₄ (÷3->0.139)

Since 0.139 is less than 0.194, CuSO₄ is limiting

(a) Al is the reactant in excess

(b) Use mols of limiting reactant to find mass of Cu formed

0.4170 mols CuSO₄ x 3 mol Cu / 3 mol CuSO₄ x 63.55 g Cu / mol Cu = 26.50 g Cu formed