(a) what mass of Fe(OH)3 in grams will precipitate from this reaction mixture?

Question

you mix 25.0 ml of 0.234 M FeCl3 with 42.5 ml of 0.453 M NaOH.
(b)one of the reactants(FeCl3 or NaOH)is present in stoichiometric excess.what is the molar concentration of the excess reactant remaining in solution after Fe(OH)3 has been precipitated?
(the large numbers are found at the base of each element)

Answer 1

First let's find how many moles of each one we have

mole= M*V, so for
FeCl3 =0.234*25*10^-3= 5.85*10^-3
NaOH =0.453*42.5*10^-3 =19.25*10^-3

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

so

1mole FeCl3 reacts with 3mole NaOH
5.85*10^-3 mole react with x

Thus x =3*5.85*10^-3=17.55*10^-3 mole NaOH but we have 19.25*10^-3 mole NaOH, therefore NaOH is in excess
Therefore the amount of Fe(OH)3 will be calculated according to the limiting reageant FeCl3.

so from the reaction

1 mole FeCl3 gives 1 mole Fe(OH)3
or better yet, since the molecular weight of Fe(OH)3 is 106.85
1 mole FeCl3 gives 106.85 g Fe(OH)3
5.85*10^-3 mole give x g
x= 106.85*5.85*10^-3= 0.625 g

We calculated that 17.55*10^-3 mole NaOH reacted.
So we have remaining (19.25-17.55) *10^-3= 1.7*10^-3 mole NaOH

These are in a total volume of 25.0+42.5=67.5 ml= 67.5*10^-3L.

So the concentration of the excess NaOH is
C=mole/V= 1.7*10^-3/67.5*10^-3= 0.025 M