Question

22. What is the theoretical yield (in g of precipitate) when 15.8 mL of a 0.642 M solution of iron(III) chloride is combined with 19.5 mL of a 0.526 M solution of lead(II) nitrate?

Answer 1

2FeCl3_(aq) + 3Pb(NO3)2_(aq) -----------> 3PbCl2_(s) + 2Fe(NO3)3_(aq)  

moles of FeCl3 = 0.642 x 15.8 /1000 = 0.0101

moles of Pb(NO3)2 = 0.526 x 19.5/1000 = 0.0102

according to balanced reaction

2 moles FeCl3 reacts with 3 moles of Pb(NO3)2

0.0101 moles of FeCl3 reacts with 0.0101 x 3 / 2 = 0.015 moles Pb(NO3)2

but we have only 0.0102 moles Pb(NO3)2. so Pb(NO3)2 is limiting reagent.

3 moles Pb(NO3)2 forms 3 moles PbCl2

0.0102 moles forms 0.0102 x 3 / 3 = 0.0102 moles PbCl2

mass = 0.0102 x 278.1 = 2.84 g

therotical yield of PbCl2 = 2.84 g