Question

What is the experimental yield (in g of precipitate) when 16 mL of a 0.5 M solution of iron(III) chloride is combined with 17.1 mL of a 0.595 M solution of silver nitrate at a 76.3% yield?

Answer 1

Ans :

The reaction is given as :

FeCl3 + 3AgNO3 = 3AgCl + Fe(NO3)3

Number of moles of FeCl3 = 0.5 x 0.016 = 0.008 mol

Number of moles of silver nitrate = 0.595 x 0.0171 = 0.0101745

3 mol AgNO3 needs 1 mol FeCl3

So 0.008 mol FeCl3 will need 3 x 0.008 = 0.024 mol AgNO3

This indicates that AgNO3 is the limiting reagent

AgCl is the solid precipitate here

Number of moles of AgCl formed = 0.0101745 mol

theoretical yield = mol x molar mass

= 0.0101745 x 143.3212

= 1.458 grams

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

76.3 = ( m / 1.458) x 100

m = 1.11 grams

So the experimental yield in g of the precipitate will be 1.11 grams.