Problem: You may want to reference (Pages 152 - 157)Section 4.4 while completing this problem.  A 57.5mL sample of a 0.112 M potassium sulfate solution is mixed with 35.0mL of a 0.110 M lead(II) acetate solution and the following precipitation reaction occurs: K2SO4(aq) + Pb(C2H3O2)2(aq) → 2 KC2H3O2(aq) + PbSO4(s) The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g.  Determine the theore©tical yield.

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Balanced equation: K2SO4(aq) + Pb(C2H3O2)2(aq) → 2 KC2H3O2(aq) + PbSO4(s) 



Recall that molarity is the ratio of the moles of solute and the volume of solution (in liters)


Molarity (M) =moles of soluteLiters of solution


Determine the limiting reactant, which is the reactant that forms the less amount of product

  • determines the maximum moles of product  theoretical yield (PbSO4)


mole-to-mole comparison 

  • 1 mole K2SOforms 1 mole PbSO4
  • 1 mole Pb(C2H3O2)2 forms 1 mole PbSO4


Molar mass PbSO = 303.26 g/mol


The mass of PbSO4 formed by 57.5mL sample of a 0.112 M  K2SO4 is:


57.5 mL solution×10-3 L1 mL×0.112 mol K2SO41 L solution×1 mol PbSO41 mol K2SO4× 303.26 g PbSO41 mol PbSO4

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Problem Details

You may want to reference (Pages 152 - 157)Section 4.4 while completing this problem.  


A 57.5mL sample of a 0.112 M potassium sulfate solution is mixed with 35.0mL of a 0.110 M lead(II) acetate solution and the following precipitation reaction occurs: 

K2SO4(aq) + Pb(C2H3O2)2(aq) → 2 KC2H3O2(aq) + PbSO4(s) 

The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g.  Determine the theore©tical yield.

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