In: Chemistry
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