Answer:
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
