Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid.5.30 g of sulfuric acid and 5.30 g of lead(II) acetate are mixed.

1) Calculate the number of grams of sulfuric acid present in the mixture after the reaction is complete.

2)Calculate the number of grams of lead(II) acetate present in the mixture after the reaction is complete.

3)Calculate the number of grams of lead(II) sulfate present in the mixture after the reaction is complete.

4) Calculate the number of grams of acetic acid present in the mixture after the reaction is complete.

Answer 1

Pb(CH3COO)2 + H2SO4 ----- PbSO4 + 2CH3COOH

Molar mass of Lead Acetate = 325.29 gm/mol

Number of moles of Lead acetate = 5.30/325.29 = 0.01629 moles

Molar mass of H2SO4 = 98.079 gm/mol

Number of moles of H2SO4 = 5.30/98.079 = 0.05403moles

1) Moles of H2SO4 left = 5.30/98.079 - 5.30/325.29 = 0.03774 moles

Mass of H2SO4 left after the reaction = 0.03774 moles * 98.079 gm/mol = 3.70198 gms

2) 0 grams, since the lead acetate is the limiting reagent in the reaction

3) Number of moles of Lead sulphate = 0.01629 moles

Molar mass of PbSO4 = 303.26 gm/mol

Mass of Lead sulphate formed = 0.01629 moles * 303.26 gm/mol = 4.940 gms

4) Number of moles of acetic acid = 0.01629 * 2 = 0.03258 moles

Molar mass of acetic acid (CH3COOH) = 2 * 12 + 2 * 16 + 4 * 1 = 60 gm/mol

Mass of acetic acid formed = 0.03258 moles * 60 gm/mol = 1.9548 gms