Final answer:
To find the volume of 0.125 M Na₂SO₄ needed to react with excess Pb(NO₃)2 and obtain 0.750 grams of solid PbSO₄, we use the balanced chemical equation and stoichiometry. The volume needed is approximately 19.8 mL.
Step-by-step explanation:
To find the volume of 0.125 M Na₂SO₄ needed to react with excess Pb(NO₃)2 and obtain 0.750 grams of solid PbSO₄, we need to consider the balanced chemical equation:
Pb(NO₃)2 (aq) + Na₂SO₄(aq) → PbSO₄ (s) + 2NaNO₃(aq)
From the equation, we can see that one mole of Pb(NO₃)2 reacts with one mole of Na₂SO₄ to produce one mole of PbSO₄. The molar mass of PbSO₄ is 303.3 g/mol. Therefore, we can calculate the number of moles of PbSO₄ using the given mass:
moles of PbSO₄ = mass / molar mass = 0.750 g / 303.3 g/mol = 0.00247 mol
Since the stoichiometry of the reaction is 1:1, we can conclude that 0.00247 mol of Na₂SO₄ is required. To find the volume, we can use the molarity equation:
molarity = moles / volume (in L)
0.125 M = 0.00247 mol / volume
Volume = 0.00247 mol / 0.125 M ≈ 0.0198 L ≈ 19.8 mL