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