Alright so I'm frusturated as ever with this stoichiometery question.
This is what I've got so far now I just need to understand the process in getting the amount of excess reactent after te reaction...
FeCl^3 + 3NaOH -------> Fe (OH)^3 + 3NaCl
m=57.4g 45.3g 37.71g 20.63g
M= 162.19 40 106.84 58.45
n= 0.353 1.132 0.352 0.352
The limiting reagent is FeCl^3= 0.353 and excess reagent is NaOH=0.3777
So yea... any help it appreciated.
If you study the stoichiometry of the equation, you would see that 1 mol of FeCl3 would react with 3 mols of NaOH.
There are only 0.353 mols present in the solution. Thus it would react only with 3 x 0.353 mols = 1.059 moles
But in the solution you have 1.132 moles.
Therefore excess NaOH is = 1.132 -1.059
= 0.073 mol
The answer you have mentioned is incorrect.