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.

### 1 Answer | Add Yours

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.

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