Answer to Question #328804 in Physical Chemistry for bean

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 = PbSO4 + 2NaNO3;

V(Pb(NO3)2) = 1.25 L;

C(Pb(NO3)2) = 0.0500 M;

V(Na2SO4) = 2.00 L;

C(Na2SO4) = 0.0250 M;

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

n(Na2SO4) = C(Na2SO4) * V(Na2SO4) = 0.0250 * 2.00 = 0.0500 M;

So, Na2SO4 is the limiting reagent and Pb(NO3)2 is the excess reagent.

n(PbSO4) = n(Na2SO4) = 0.0500 M;

M(PbSO4) = 303 g/mol;

m(PbSO4) = n(PbSO4) * M(PbSO4) = 0.0500 * 303 = 15.15 g.

Answer. The limiting reagent is Na2SO4;

mass of PbSO4 is 15.15 g.