Answer:
The answer to your question is the limiting reactant is CuSOâ‚„ and 0.975 moles of Cu were obtained
Explanation:
moles of Copper = ?
mass of Aluminum = 29 g
mass of CuSOâ‚„ = 156 g
Limiting reactant = ?
Balanced Chemical reaction
         3 CuSO₄  +  2 Al  ⇒  3 Cu  +  Al₂(SO₄)₃
Calculate the moles of reactants
CuSOâ‚„ = 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 CuSOâ‚„. Â Â Â
          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.    Â
Answer:
0.977 moles of Copper are obtained. The limiting reagent is Copper (II) Sulfate.
Explanation:
First you need to write out a balanced chemical equation. Then you convert from 29.0 grams of Aluminum to moles of Aluminum. Take moles of Aluminum and convert to moles of Copper to get 1.61 moles. Then you take 156 grams of Copper (II) Sulfate and convert it to moles of Copper (II) Sulfate. After you have the moles of Copper (II) Sulfate you then need to convert it to moles of Copper to get 0.977 moles of Copper. 0.977 is the smaller number so it will be your answer and Copper (II) Sulfate will be your limiting reagent. Remember to also use significant figures when required.
