A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) = Al2(SO4)3(aq) + 3Cu(s)
Select one:
a)Copper
b)Aluminum sulfate
c)Aluminum
d)Copper (II) sulfate

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.

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

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