3cuso4 + 2al ------ 3cu + al2(so4)3 in the limiting reagent question above if you have 8.3 g of cuso4 and 20.1 g of al how many grams of copper can you produce?
balanced equation for the reaction is; 3CuSO₄ + 2Al ---> Al₂(SO₄)₃ + 3Cu Stoichiometry of CuSO₄ to Al is 3:2 Number of CuSO₄ moles present - 8.3 g/ 159.6 g/mol = 0.052 mol Number of Al moles present - 20.1 g/ 27 g/mol = 0.74 mol We need to first find the limiting reagent. Limiting reagent is the reactant that is fully consumed, amount of product formed depends on the amount of limiting reactant present. 3 mol of CuSO₄ reacts with 2 mol of Al Therefore 0.052 mol of CuSO₄ reacts with - 2/3 x 0.052 = 0.034 mol of Al required. But 0.74 mol of Al present , therefore Al is in excess and CuSO₄ is limiting reactant, stoichiometry of CuSO₄ to Cu is 3:3 = 1:1 Therefore number of Copper moles formed - 0.052 mol Then mass of Copper formed - 0.052 mol x 63.5 g/mol = 3.30 g of Copper is produced