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

mass PbSO4=5.50 g H2SO4×1 mol H2SO498.079 g H2SO4                                ×1 mol PbSO41 mol H2SO4×303.26 g PbSO41 mol PbSO4

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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.

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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