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How much of the excess reactant remains after the reaction?Alright so I'm frusturated...

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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.

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Posted (Answer #2)

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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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