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artcher [175]
3 years ago
12

Consider the dissolution equation of lead(II) chloride.

Chemistry
1 answer:
Vladimir79 [104]3 years ago
4 0

Answer:

The solubility product of lead(II) chloride is 1.61\times 10^{-5}.

Explanation:

Concentration of lead (II) ions = [Pb^{2+}]=0.0159 M

Concentration of chloride ion = [Cl^-]=0.0318 M

PbCl_2(s)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

The expression of a solubility product will be given as:

K_{sp}=[Pb^{2+}][Cl^-]^2

=0.0159 M\times (0.0318 M)^2=1.61\times 10^{-5}

The solubility product of lead(II) chloride is 1.61\times 10^{-5}.

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

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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.

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