# 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.

###### FREE Expert Solution

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)

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:

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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.

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Solution Stoichiometry concept. You can view video lessons to learn Solution Stoichiometry. Or if you need more Solution Stoichiometry practice, you can also practice Solution Stoichiometry practice problems.