Question

You mix 28.0 mL of 0.298 M FeCl₃ with 39.5 mL of 0.423 M NaOH. What mass of Fe(OH)₃ (in grams) will precipitate from this reaction mixture?

One of the reactants (FeCl₃ or NaOH) is present in a stoichiometric excess. What is the molar concentration of the excess reactant remaining in solution after Fe(OH)₃ has been precipitated?

Answer 1

First the balanced reaction is

FeCl3 (aq) + 3 NaOH ( aq) ----> Fe(OH)3 (s) +   3 NaCl (aq)

Now we find moles of NaoH and moles of FeCL3

NaOH moles = M x V ( in liters) = ( 0.423 x 39.5 /1000) = 0.0167085

FeCl3 moles = M x V ( in liters) = 0.298 x ( 28/1000) = 0.008344

As per reaction we see that 1 FeCl3 reacts with 3NaOH

Hnece for 0.008344 moles FeCl3 we need NaOH moles = 3 x 0.008344 = 0.025032

But we had only 0.0167085 moles NaOH . Hence NaOH is limiting reagent while FeCl3 is excess reagent.

Product Fe(OH)3 depends on limiting regaent NaOH moles.

Fe(OH)3 moles = ( 1/3) NaoH moles = ( 1/3) x 0.0167085 = 0.0055695

Fe(OH)3 mass = moles x molar mass of Fe(OH)3 = 0.0055695 x 106.867 = 0.595 g

Now we fidn excess FeCl3.

FeCL3 moles reacted = ( 1/3) NaOH moles =   0.0055695

Now FeCl3 moles left unreacted = initial FeCL3 moles = FeCL3 moles reacted = 0.025032 - 0.0055695

            = 0.0194625

volume of solution = 28+39.5 = 67.5 ml = 0.0675 L

Now excess [FeCl3] = moles/vol in L = 0.0194625 / 0.0675 = 0.288 M