Problem 80 - Chemistry The Central Science

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.

Step-by-Step Solution

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 1: 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: \( H_2SO_4 + Pb(C_2H_3O_2)_2 \rightarrow PbSO_4 + 2CH_3COOH \)

Step 2: 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

Step 3: 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.

Step 4: 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

Step 5: 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.