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Chapter 3: Problem 76

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. If \(7.50 \mathrm{~g}\) of sulfuric acid and \(7.50 \mathrm{~g}\) 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

Expert verified
After the reaction is complete, there are 5.28 g of sulfuric acid remaining, 0 g of lead(II) acetate remaining (reacted completely), 6.99 g of lead(II) sulfate formed, and 2.77 g of acetic acid formed.

Step by step solution

01

Write the balanced chemical equation

First, we should write out the unbalanced chemical equation and identify each component: Sulfuric acid (H2SO4) + Lead(II) acetate (Pb(C2H3O2)2) 鉄� Lead(II) sulfate (PbSO4) + Acetic acid (HC2H3O2) To balance the equation, we need equal numbers of each type of atom on each side: H2SO4 + Pb(C2H3O2)2 鉄� PbSO4 + 2 HC2H3O2
02

Determine the molar mass of each reactant and product

To convert between grams and moles, we need to find the molar mass of each substance: - Molar mass of H2SO4 = (2脳1.008) + 32.07 + (4脳16.00) = 98.08 g/mol - Molar mass of Pb(C2H3O2)2 = 207.2 + 2(2脳12.01 + 3脳1.008 + 2脳16.00) = 325.21 g/mol - Molar mass of PbSO4 = 207.2 + 32.07 + 4脳16.00 = 303.26 g/mol - Molar mass of HC2H3O2 = 1.008 + 2脳12.01 + 3脳1.008 + 2脳16.00 = 60.05 g/mol
03

Convert the given masses of reactants to moles

Given the mass (in grams) of each reactant and their molar masses, we can calculate the number of moles for each reactant: Moles of H2SO4 = (7.50 g) / (98.08 g/mol) = 0.0765 mol Moles of Pb(C2H3O2)2 = (7.50 g) / (325.21 g/mol) = 0.0231 mol
04

Determine the limiting reactant

Using the balanced chemical equation, we can determine the ratio of moles of H2SO4 to Pb(C2H3O2)2 needed for the reaction: H2SO4 : Pb(C2H3O2)2 = 1 : 1 To determine the limiting reactant, we compare the mole ratios of available reactants: 0.0765 mol H2SO4 / 1 = 0.0765 0.0231 mol Pb(C2H3O2)2 / 1 = 0.0231 Since 0.0231 < 0.0765, Pb(C2H3O2)2 is the limiting reactant.
05

Find the grams of products formed and reactants remaining

Using stoichiometry, we can calculate the grams of lead(II) sulfate and acetic acid formed and the grams of sulfuric acid remaining: Pb(C2H3O2)2 (limiting reactant) 鉄� PbSO4 (product) 0.0231 mol x (303.26 g/mol) = 6.99 g of PbSO4 Pb(C2H3O2)2 (limiting reactant) 鉄� 2 HC2H3O2 (product) 0.0231 mol x 2 脳 (60.05 g/mol) = 2.77 g of HC2H3O2 7.50 g of H2SO4 (initial) - [0.0231 mol x (98.08 g/mol)] = 5.28 g of H2SO4 (remaining)
06

Report the final grams of each substance

After the reaction is complete, the amounts of each substance are as follows: - Sulfuric acid (H2SO4): 5.28 g remaining - Lead(II) acetate (Pb(C2H3O2)2): 0 g remaining (reacted completely) - Lead(II) sulfate (PbSO4): 6.99 g formed - Acetic acid (HC2H3O2): 2.77 g formed

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Most popular questions from this chapter

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