# How much of the excess reactant remains after the reaction?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...

How much of the excess reactant remains after the reaction?

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.

thilina-g | Certified Educator

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.