# Problem: Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 5.50 g of sulfuric acid and 5.50 g of lead(II) acetate are mixed.Calculate the number of grams of sulfuric acid present in the mixture after the reaction is complete.

###### FREE Expert Solution
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###### FREE Expert Solution

The chemical reaction is:

H2SO4(aq) + Pb(C2H3O2)2(aq) → PbSO4(s) + HC2H3O2(aq)

Balancing this gives us:

H2SO4(aq) + Pb(C2H3O2)2(aq) → PbSO4(s) + 2 HC2H3O2(aq)

We're being asked to determine the mass of H2SO4 present in the mixture after the reaction is complete. This means that H2SO4 is the excess reactant while Pb(C2H3O2)2 is the limiting reactant.

The molar mass of H2SO4 is 2(1.01 g/mol H) + 32.06 g/mol S + 4(16.00 g/mol O) = 98.08 g/mol. The molar mass of Pb(C2H3O2)2 is 207.20 g/mol Pb + 4(12.01 g/mol C) + 6(1.01 g/mol H) + 4(16.00 g/mol O) = 325.30 g/mol. From the balanced chemical equation, 1 mole of Pb(C2H3O2)2 reacts with 1 mole of H2SO4.

81% (300 ratings)
###### Problem Details
Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 5.50 g of sulfuric acid and 5.50 g of lead(II) acetate are mixed.

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