Question

What is the experimental yield (in g of precipitate) when 16.3 mL of a 0.7 M solution of iron(III) chloride is combined with 15.9 mL of a 0.704 M solution of lead(II) nitrate at a 88.7% yield?

Answer 1

Ans :

The reaction is given as :

2FeCl3 + 3Pb(NO3)2 = 2Fe(NO3)3 + 3PbCl2

Number of mol of FeCl3 : 0.7 x 0.0163 = 0.01141 mol

Number of mol of Pb(NO3)2 = 0.704 x 0.0159 = 0.0111936

2 mol FeCl3 need 3 mol Pb(NO3)3 , so lead (II) nitrate is the limiting reagent here

PbCl2 is the solid precipitate formed here.

Number of moles of PbCl2 formed will be : 0.0111936 mol

So theoretical yield = mol x molar mass

= 0.0111936 x 278.106

= 3.113 grams

Percent yield = ( experimental yield / theoretical yield) x 100

88.7 = ( m / 3.113) x 100

m = 2.76 g

So the experimental yield of solid precipitate in grams will be 2.76 g.