Question

What is the experimental yield (in g of precipitate) when 18.8 mL of a 0.5 M solution of iron(III) chloride is combined with 15.4 mL of a 0.691 M solution of silver nitrate at a 76.8% yield?

Answer 1

Number of moles,n= Molarity*volume in L

nFeCl3 = 0.5M*18.8 mL*10^-3L/mL

= 0.0094 mol

nAgNO3= 0.691M*15.4 mL*10^-3L/mL

= 0.0106 mol

The balanced equation is

FeCl3(aq)+3AgNO3(aq)---> Fe(NO3)3(aq)+3AgCl(s)

3 moles of AgNO3 reacts with 1 mole of Fe(NO3)3

0.0106 moles of AgNO3 reacts with (0.0106/3)=0.0035 moles of Fe(NO3)3

So 0.0094-0.0035= 0.0059 moles of Fe(NO3)3 left unreacted and is excess reactant.

Hence AgNO3 is the limiting reactant.

From the balanced equation

3 moles of AgNO3 produces 3 moles of AgCl

0.0106 moles of AgNO3 produces 0.0106 moles of AgCl

Mass of AgCl = number of moles*molarmass

= 0.0106 mol* 143.3 g/mol

= 1.519 g

For 100% the yield is 1.519 g

For 76.8 % the yield is (76.8/100)*1.519= 1.167g