Question

What is the experimental yield (in g of precipitate) when 16.6 mL of a 0.5 M solution of iron(III) chloride is combined with 17.2 mL of a 0.567 M solution of lead(II) nitrate at a 89.8% yield

Answer 1

the reaction is

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

Thus 2 moles of Fecl3 produces 2 moles of PbCl2

We need to fing the limiting reagent first.

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

16.6x0.5=8.3 17.2x0.567= 9.7524 0 0

8.3/2 =4.15 9.7524/3=3.25

As the ratio of lead nitrate is less, it is the limiting reagent

Thus the product is to be calculated from it.

Thus 3 moles of lead nitrate produces 2 moles of lead chloride

9.7524x10^-3 mol gives = 9.7524x10^-3 mol x2mol/3mol

= 6.5016x10^-3 mol

Thus theoretical yield of PbCL2 = moles x molar mass

= 6.5016 x10^-3mol x278.1 g/mol

=1.808 g

If the experimental yield was 89.8% then

the mass of precipitate = 1.808 x 89.8/100

= 1.62 g

The experimental yield was 1.62 g