Question

When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed.

Answer 1

Answer:

The mass of PbSO4 formed 15.163 gram

Explanation:

mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625

mole of Na₂SO₄ = 2 x 0.025 = 0.05

                                      Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃

( Mole/Stoichiometry )    $\frac{0.0625}{1}$           $\frac{0.05}{1}$

                                     = 0.0625     = 0.05

From  (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.

Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄

                                            = 0.05 x 303.26 g

                                            = 15.163 g