Question

Copper is formed when aluminum reacts with copper (II) sulfate in a single-replacement reaction. How many moles of copper can be obtained when 29.0 g of Al reacts with 156 g of copper (II) sulfate? What is the limiting reagent?

Answer 1

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.

Answer 2

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.