Answer to Question #328834 in General Chemistry for poster

Question

When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the limiting reagent and mass of PbSO4 when a 1.25 L solution of 0.0500 M Pb(NO3)2 and a 2.00 L solution of 0.0250 M Na2SO4 are mixed. (Hint: start first by balancing the equation of the reaction between Na2SO4 and Pb(NO3)2

Answer 1

Na₂SO₄ + Pb(NO₃)₂ = 2NaNO₃ + PbSO₄

n(Na₂SO₄) = C(Na₂SO₄) * V(Na₂SO₄) = 0.0250 M * 2.00 L = 0.05 moles

n(Pb(NO₃)₂) = C(Pb(NO₃)₂) * V(Pb(NO₃)₂) =  0.0500 M * 1.25 L = 0.0625 moles

Pb(NO₃)₂ excess reagent

Na₂SO₄ limiting reagent

m(PbSO₄) = (m(Na₂SO₄)*Mr(PbSO₄)) / Mr(Na₂SO₄) = n(Na₂SO₄)*Mr(PbSO₄)

m(PbSO₄) = 0.05 moles * 303 g/mol = 15.15 g