Question

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant?

Answer 1

Answer:

d. Copper (II) sulfate

Explanation:

Given data:

Mass of Al = 1.25 g

Mass of CuSO₄ = 3.28 g

What is limiting reactant = ?

Solution:

Chemical equation:

2Al + 3CuSO₄   →   Al₂ (SO₄)₃ + 3Cu

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 1.25 g/ 27 g/mol

Number of moles = 0.05 mol

Number of moles of CuSO₄:

Number of moles = mass/molar mass

Number of moles = 3.28 g/ 159.6 g/mol

Number of moles = 0.02 mol

now we will compare the moles of reactant with product.

               Al           :           Al₂ (SO₄)₃

                 2          :             1

               0.05       :          1/2×0.05=0.025 mol

                Al           :            Cu

                 2            :              3

               0.05         :            3/2×0.05 = 0.075 mol

         CuSO₄           :           Al₂ (SO₄)₃

                3             :             1

               0.02         :          1/3×0.02=0.007 mol

         CuSO₄           :            Cu

               3               :              3

               0.02         :              0.02

Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.