If 100.0 mL of 0.105 M Na 2 SO 4 are added to 100.0 mL of 0.985 M Pb (NO 3) 2, how many grams of PbSO 4 can be produced? Na 2 SO 4 (aq) + Pb (NO 3) 2 (aq) ⟶ 2 NaNO 3 (aq) + PbSO 4 (s)

3.18 g of lead sulfate is produced in this reaction

Explanation:

The reaction is this one:

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

Let's determine the moles of reactants, to find out which is the limiting.

Molarity. volume = Mol

0.105 m/L. 0.100L = 0.0105 moles of sulfate

0.985 m/L. 0.100L = 0.0985 moles of nitrate

Ratio is 1:1, so 1 mol of sulfate needs 1 mol of nitrate, to react.

0.0985 moles of nitrate need 0.0985 moles of sulfate to react, but I only have 0.0105 moles so the sulfate is my limiting reactant.

Ratio is again 1:1, so 0.0105 moles of sulfate make 0.0105 moles of lead sulfate.

Mol. molar mass = mass → 0.0105 mol. 303.26 g/m = 3.18 g