Answer:
The answer to your question is the limiting reactant is CuSO₄ and 0.975 moles of Cu were obtained
Explanation:
moles of Copper = ?
mass of Aluminum = 29 g
mass of CuSO₄ = 156 g
Limiting reactant = ?
Balanced Chemical reaction
3 CuSO₄ + 2 Al ⇒ 3 Cu + Al₂(SO₄)₃
Calculate the moles of reactants
CuSO₄ = 64 + 32 + (16 x 4) = 160g
Al = 27 g
160 g of CuSO₄ ----------------- 1 mol
156 g ----------------- x
x = (156 x 1) / 160
x = 0.975 moles
27 g of Al -------------------------- 1 mol
29 g of Al -------------------------- x
x = (29 x 1)/27
x = 1.07 moles
Calculate proportions to find the limiting reactant
Theoretical 3 moles CuSO₄/2 moles Al = 1.5 moles
Experimental 0.975 moles CuSO₄/1.07 moles = 0.91
The experimental proportion was lower than the theoretical proportion that means that the limiting reactant is CuSO₄.
3 moles of CuSO₄ ------------------ 3 moles of Cu
0.975 moles of CuSO₄ --------------- x
x = (0.975 x 3)/3
x = 0.975 moles of Cu were obtained.
