Question

Exercise 4.70 - Enhanced - with Feedback 《" 14of14 You may want to reference (Pages 152- 157) Section 4.4 while completing this problem. A 56.0 mL sample of a 0.106 M potassium sultate solution is mixed with 35.0 mL. of a 0.112 M lead(II) acetate solution and the following precipitation reaction oocurs: K2S04(aq) + Pb(C2Hs02)2(aq)+2KC2HsO2(aq) + Pbso4(s) The solid PbSO4 is collected, dried, and found to have a mass of 0 999 g Determine the limiting reactant, the theoretical yield, and the percent yield. Part A

Answer 1

Number of moles of Potassium Sulfate (K2SO4) = Volume of solution (in L) * Molarity (M) = 56.0/1000 * 0.106 = 0.005936 moles

Number of moles of Lead(II) acetate = Volume of solution (in L) * Molarity (M) = 35.0/1000 * 0.112 = 0.00392 moles

Since K2SO4 and Lead(II) acetate reacts in 1:1 ratio

Hence the limiting reagent will be Lead (II) acetate

One mole of Lead (II) acetate gives one mole of PbSO4

Moles of PbSO4 formed (theoritical yield) = 0.00392 moles

Molar mass of PbSO4 = 303 gm/mol

Theoritical yield in grams = Number of moles * Molar mass = 0.00392 mol * 303 gm/mol = 1.19 grams

Percent Yield = Actual Yield/Theoritical Yield * 100 = 0.999/1.19 * 100 = 83.9%