When aqueous solutions of Na2SO4 and Pb (NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.35L of 0.125 M Pb(NO3)2 and 3.00L of 0.0250M Na2SO4 are mixed.

We have given,

Volume of pb(NO₃)₂ = 1.35 L

Concentration of pb(NO3)2 = 0.125 M

Volume of Na₂SO₄ = 3.00 L

Concentration of Na₂SO₄ = 0.00250 M

Number of moles of pb (NO₃)₂ = 1.35 L x 0.125 M

= 0.16875 mole

or, 0.169 mol.

Number of moles of Na₂SO₄ = 3.00 L x 0.0250 M

= 0.075 mol

Na₂SO₄ + Pb(NO₃)₂ → 2NaNO₃ + PbSO₄ ↓

∵ One mole of Na₂SO₄ reacts with one mole of Pb(NO₃)₂ and forms,

Here,

Number of moles of Na₂SO₄ is less than the number of moles of Pb(NO₃)₂. 

It means, 

Na₂SO₄ is a limiting reagent.

∴ Number of moles of PbSO₄ formed = 0.075 mol

Molar mass of PbSO₄ = 303.26 g/mol

∴ Mass of PbSO₄ = 0.075 mol x 303.26 g/mol

= 22.74 g

Hence,

Mass of PbSO₄ formed = 22.74 g