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.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.

FREE Expert Solution

Reactants:

sulfuric acid → H₂SO₄
lead (II) acetate → Pb(CH₃COO)₂

Products:

lead (II) sulfate → PbSO₄
acetic acid → CH₃COOH

Balanced Reaction: H₂SO₄(aq) + Pb(CH₃COO)₂(aq) → PbSO₄(s) + 2 CH₃COOH(aq)

• PbSO₄(s) from H₂SO₄(aq)

molar mass PbSO₄ = 303.26 g/mol
molae mass H₂SO₄ = 98.079 g/mol

$$\begin{gathered} \mathbf{mass}\mathbf{ }{\mathbf{PbSO}}_{\mathbf{4}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{50}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}}{\mathbf{98}\mathbf{.}\mathbf{079}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{PbSO}}_{\mathbf{4}}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}}\mathbf{\times }\frac{\mathbf{303}\mathbf{.}\mathbf{26}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{PbSO}}_{\mathbf{4}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{PbSO}}_{\mathbf{4}}}} \end{gathered}$$

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