chapter 9 Exercise 938 with feedback tExercise 9.38 with feedback e previous 4 o
ID: 513246 • Letter: C
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chapter 9 Exercise 938 with feedback tExercise 9.38 with feedback e previous 4 of 24 next, You may want to reference (D -pages 301.308) Section 9.3 while c A 60.5 ompleting this problem. mL sample of a 0.122 M potassium sulfate solution is mixed with 37 0 mL of a 0.128 Miead(II acetate solution and the following preciptation reaction occurs: The solid PbSO, is collected, dried, and found to have a mass of 1.01 g Determine the limiting reactant, the theoretical yield, and the percent yield Part A Identify the limiting reactant. O KC2H30 O PbSO. O K2SO4 Submit My Answers GheUR Part B Determine the theoretical yield.Explanation / Answer
The baanced equation is
K2SO4(aq) + Pb(C2H3O2)2 (aq) -------------> PbSO4(s) + KC2H3O2 (aq)
60.5 x 0.122 37x0.128 0 0 initial mmoles
= 7.381 =4.736
Thus lead acetate is the limiting reagent.
According to balanced equation 1 mole of lead acetate give 1 mole of lead sulphat.
Thus 4.736 mmoles of lead acetate should give 4.736 mmoles of lead sulphate.
hence
mmoles of lead sulphate to be formed = 4.736 mmoles
b)Theoretical yiled of lead sulphate = moles x molar mass
= 4.736 x 10-3 x303.26g/mol
=1.436g
c) percentage yield of lead sulphate = experimental yield x 100/ theoretical yield
= 1.01g x100 /1.436 g
= 70.33 %
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