Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. If \(5.00 \mathrm{~g}\) of sulfuric acid and \(5.00 \mathrm{~g}\) of lead(II) acetate are mixed, calculate the number of grams of sulfuric acid, lead(II) acetate, lead(II) sulfate, and acetic acid present in the mixture after the reaction is complete.

Short answer

After the reaction is complete, 3.49 g of sulfuric acid remains, all 5.00 g of lead(II) acetate is consumed, 4.67 g of lead(II) sulfate is formed, and 1.85 g of acetic acid is formed.

Step-by-step solution

Write down the balanced chemical equation

The balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and lead(II) acetate (Pb(C₂H₃O₂)₂) is: \( H_2SO_4 + Pb(C_2H_3O_2)_2 \rightarrow PbSO_4 + 2CH_3COOH \)

Convert the given masses of reactants to moles

Let's first find the molar masses of the reactants: - H₂SO₄: 2(1.01) + 32.07 + 4(16.00) = 98.08 g/mol - Pb(C₂H₃O₂)₂: 207.2 + 2(2(12.01) + 3(1.01) + 16.00) = 325.3 g/mol Now, convert the given masses (5.00 g each) of reactants into moles: Moles of H₂SO₄ = \(\frac{5.00\: \mathrm{g}}{98.08\: \mathrm{g/mol}}\) = 0.0510 mol Moles of Pb(C₂H₃O₂)₂ = \(\frac{5.00\: \mathrm{g}}{325.3\: \mathrm{g/mol}}\) = 0.0154 mol

Determine the limiting reactant

From the balanced equation, the stoichiometry between H₂SO₄ and Pb(C₂H₃O₂)₂ is 1:1. We can find the limiting reactant by comparing the mole ratios: Mole ratio H₂SO₄:Pb(C₂H₃O₂)₂ = 0.0510:0.0154 = 3.31:1.00 Since the stoichiometry requires a 1:1 ratio, Pb(C₂H₃O₂)₂ is the limiting reactant.

Calculate the moles of products formed

Using the stoichiometry from the balanced equation, we can calculate the moles of products formed: Moles of PbSO₄ = Moles of Pb(C₂H₃O₂)₂ = 0.0154 mol Moles of CH₃COOH = 2 x Moles of Pb(C₂H₃O₂)₂ = 2 x 0.0154 = 0.0308 mol

Convert the moles of products back to grams

Determine the molar masses of the products: - PbSO₄: 207.2 + 32.07 + 4(16.00) = 303.27 g/mol - CH₃COOH: 12.01 + 4(1.01) + 16.00 = 60.05 g/mol Now, convert the moles of products into grams: Mass of PbSO₄ = 0.0154 mol x 303.27 g/mol = 4.67 g Mass of CH₃COOH = 0.0308 mol x 60.05 g/mol = 1.85 g The remaining amount of H₂SO₄ can also be determined since we know the mass in the beginning and the moles that reacted: Mass of remaining H₂SO₄ = Initial moles of H₂SO₄ – moles that reacted = 0.0510 mol – 0.0154 mol = 0.0356 mol Mass of remaining H₂SO₄ = 0.0356 mol x 98.08 g/mol = 3.49 g After the reaction is complete: - 3.49 g of sulfuric acid remains. - All 5.00 g of lead(II) acetate is consumed. - 4.67 g of lead(II) sulfate is formed. - 1.85 g of acetic acid is formed.