Solutions of sulfuric acid and lead (ll) acetate react to form solid lead (ll) sulfate and a solution of acetic acid.?

Molar mass of H₂SO₄ = (1.0×2 + 32.1 + 16.0×4) g/mol = 98.1 g/mol
Molar mass of Pb(CH₃COO)₂ = (207.2 + 12.0×4 + 1.0×6 + 16.0×4) g/mol = 325.2 g/mol
Molar mass of PbSO₄ = (207.2 + 32.1 + 16.0×4) g/mol = 303.3 g/mol
Molar mass of CH₃COOH = (12.0×2 + 1.0×4 + 16.0×2) g/mol = 60.0 g/mol

Initial number of moles of H₂SO₄ = (10.0 g) / (98.1 g/mol) = 0.10194 mol
Initial number of moles of (CH₃COO)₂Pb = (10.0 g) / (325.2 g/mol) = 0.03075 mol

H₂SO₄(aq) + Pb(CH₃COO)₂(s) → PbSO₄(aq) + 2CH₃COOH(aq)
Mole ratio H₂SO₄(aq) : Pb(CH₃COO)₂ : PbSO₄ : CH₃COOH = 1 : 1 : 1 : 2

When 0.03075 mol Pb(CH₃COO)₂ completely reacts, H₂SO₄ needed = 0.03075 mol < 0.10194 mol
Hence, H₂SO₄ is excess, and Pb(CH₃COO)₂ is the limiting reagent/reactant.
Number of moles of Pb(CH₃COO)₂ reacted = 0.03075 mol
Number of moles of H₂SO₄ reacted = 0.03075 mol
Number of moles of PbSO₄ produced = 0.03075 mol
Number of moles of CH₃COOH produced = (0.03075 mol) × 2 = 0.0615 mol

After the reaction is complete:
Mass of Pb(CH₃COO)₂ present in the reaction mixture = 0 g (completely reacted)
Mass of H₂SO₄ present in the reaction mixture = [(0.10194 - 0.03075) mol] × (98.1 g/mol) = 6.98 g
Mass of PbSO₄ present in the reaction mixture = (0.03075 mol) × (303.3 g/mol) = 9.33 g
Mass of CH₃COOH present in the reaction mixture = (0.0615 mol) × (60.0 g/mol) = 3.69 g