Solution Stoichiometry

98) m = 1.50 g Pb(NO₃)₂          [Na₂SO₄] = 0.100 M          V = 125 mL

      (a) Pb(NO₃)₂(aq) + Na₂SO₄(aq) 2NaNO₃(aq) + PbSO₄(s)

            Pb²⁺(aq) + SO₄^2-(aq) PbSO₄(s)

      (b) [SO₄^2-] = n/V

            n_SO42- = 0.100 mol Na₂SO₄/L x 125 mL x 1 L/10³ mL x 1 mol SO₄^2-/1 mol Na₂SO₄ = 1.25 x 10^-2 mol SO₄^2-

            n_Pb2+ = 1.50 g Pb(NO₃)₂ x 1 mol Pb(NO₃)₂/331.22 g Pb(NO₃)₂ x 1 mol Pb ²⁺/1 mol Pb(NO₃)₂ = 4.53 x 10^-3 mol Pb²⁺

            n_PbSO4 = 1.25 x 10^-2 mol SO₄^2- x 1 mol PbSO₄/1 mol SO₄^2- = 1.25 x 10^-2 mol PbSO₄

            n_PbSO4 = 4.53 x 10^-3 mol Pb²⁺ x 1 mol PbSO₄/1 mol Pb²⁺ = 4.53 x 10^-3 mol PbSO₄

            The limiting reactant produces the smallest amount of product, therefore the Pb(NO₃)₂ is the limiting reactant.

      (c) Pb²⁺(aq) + 2NO₃^-(aq) + 2Na⁺(aq) + SO₄^2-(aq) 2Na⁺(aq) + 2NO₃^-(aq) + PbSO₄(s)

           [Pb²⁺] = n/V = 0

           [NO₃^-] = 4.53 x 10^-3 mol Pb²⁺ x 1 mol Pb(NO₃)₂/1 mol Pb²⁺ x 2 mol NO₃^-/1 mol Pb(NO₃)₂/(125 mL x 1 L/10³ mL)

           [NO₃^-] = 7.25 x 10^-2 M

           [Na⁺] = 0.100 mol Na₂SO₄/L x 2 mol Na⁺/ 1 mol Na₂SO₄ = 0.200 M

           n_SO42-(start) = 0.100 mol Na₂SO₄/L x 125 mL x 1 L/10³ mL = 0.0125 mol SO₄^2-

           n_SO42-(reacted) = 4.53 x 10^-3 mol Pb²⁺ x 1 mol SO₄^2- /1 mol Pb²⁺ = 4.53 x 10^-3 mol SO₄^2-

           [SO₄^2-] = n_SO42-(start) - n_SO42-(reacted)/V

           [SO₄^2-] = (0.0125 mol SO₄^2- - 4.53 x 10^-3 mol SO₄^2-)/(125 mL x 1 L/10³mL) = 0.064 M