Question

What is the experimental yield (in g of precipitate) when 18.6 mL of a 0.6 M solution of iron(III) chloride is combined with 15.9 mL of a 0.738 M solution of silver nitrate at a 77.8% yield?

Answer 1

3AgNO3 +FeCl3= 3AgCl+ Fe(NO3)₃

_{143g 162.5}

moles of AgNO3=18.6*0.6/1000=0.01116 ,moles of FeCl3=15.9*0.738/1000=0.0117342

mass of AgNO3= 0.01116*143=1.5958g   mass of FeCl3 =0.0117342*162.5=1.9068g

AgNO3 is the limiting reagent

Having identified silver nitrate as the limiting reagent we can do the rest of the calculation:

0.01116 moles=0.01116*179=1.99764 g of Agcl

% yield = x/1.99764=0.778 so x=1.55416 g of AgCl precipitate is the experimental yield