Question

21. What is the experimental yield (in g of precipitate) when 16.5 mL of a 0.7 M solution of iron(III) chloride is combined with 15.8 mL of a 0.721 M solution of lead(II) nitrate at a 87.4% yield?

Answer 1

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

moles of FeCl3 = 0.7 x 16.5/1000 = 0.0115

moles of Pb(NO3)2 = 0.721 x 15.8/1000 = 0.0114

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

0.0115 moles FeCl3 reacts with 0.0115 x 3 / 2 = 0.0172

but we have only 0.0114 moles Pb(NO3)2. so Pb(NO3)2 is limiting reagent.

3 moles Pb(NO3)2 forms 3 moles PbCl2

0.0114 moles Pb(NO3)2 forms 0.0114 x 3 / 3 = 0.0114 moles

mass of PbCl2 = 0.0114 x 278.1 = 3.17 g

therotical yield = 3.17 g

% yield = 87.4

87.4 = (X / 3.17) x 100

X / 3.17 = 0.874

X = 2.77 g

experimental yield = 2.77 g