Question

What is the experimental yield (in g of precipitate) when 15.9 mL of a 0.6 M solution of iron(iii) chloride is combined with 19.9 mL of a 0.738 M solution of lead(ok) nitrate at a 76.6% yield?

Answer 1

2FeCl3 (aq) + 3Pb(NO3)2 (aq)-----------------> 3PbCl2(s) + 2Fe(NO3)3(aq)

no of moles of FeCl3   = molarity * volume in L

                                   = 0.6*0.0159 = 0.00954 moles

no of moles of Pb(NO3)2 = molarity * volume in L

                                         = 0.738*0.0199 =0.0147moles

3 moles of Pb(NO3)2 react with 2 moles of Fe(NO3)3

0.0147 moles of Pb(NO3)2 react with = 2*0.0147/3   = 0.098 moles of Fe(NO3)3

FeCl3 is limiting reactant

2 moles of FeCl3 react with Pb(NO3)2 to gives 3 moles of Pb(NO3)2

0.00954 moles of FeCl3 react with Pb(NO3)2 to gives = 3*0.00954/2 = 0.01431 moles of Pb(NO3)2

experimental yield of Pb(NO3)2 = no of moles * gram molar mass

                                              = 0.01431*331   = 4.74g

percent yield    =   experimental yield*100/theorical yield

76.6                 = experimental*100/4.74

experimental yield      = 76.6*4.74/100    = 3.63g