Reactants:

sulfuric acid → H_{2}SO_{4}

lead (II) acetate → Pb(CH_{3}COO)_{2}

lead (II) acetate → Pb(CH

Products:

lead (II) sulfate → PbSO_{4}

acetic acid → CH_{3}COOH

**Balanced Reaction: H _{2}SO_{4}(aq) + Pb(CH_{3}COO)_{2}(aq) → PbSO_{4}(s) + 2 CH_{3}COOH(aq)**

**• PbSO _{4}(s) from H_{2}SO_{4}(aq)**

molar mass PbSO_{4} = 303.26 g/mol

molae mass H_{2}SO_{4} = 98.079 g/mol

$\mathbf{mass}\mathbf{}{\mathbf{PbSO}}_{\mathbf{4}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{50}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}}{\mathbf{98}\mathbf{.}\mathbf{079}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{PbSO}}_{\mathbf{4}}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}}\mathbf{\times}\frac{\mathbf{303}\mathbf{.}\mathbf{26}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{PbSO}}_{\mathbf{4}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{PbSO}}_{\mathbf{4}}}}$

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.

You may want to reference (Pages 106 - 110)Section 3.7 while completing this problem.

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

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