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Offline xstrae

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Stoichiometry question
« on: January 01, 2007, 01:36:31 AM »
Hi,

1.44g of FeC2O4 was dissolved in sulphuric acid(dil.) and the solution was diluted to 100cm3. Which of the following solution of KMnO4 will be able to oxidise it completely? (multiple options may be correct)

(A) 0.01M , 400cm3
(B) 0.02M , 200cm3
(C) 0.01M , 600cm3
(D) 0.05N , 600cm3

My work:
For complete oxidation, Milli equivalents of FeC2O4 = Milli equivalents of KMnO4

Milli equivalents of FeC2O4 = wt/eq.wt x 1000
                                                                    = 1.44/72 x 1000  = 20
(  eq wt = m.w / n factor = 144/2  )

In acidic medium, Mn+7 ---->  Mn+2
therefore, n factor of KMnO4 = 5

N = n x M
Ml. eq of KMnO4 = N x V

Putting in the values, I got (A) and (B) as the correct answer. My book, however, lists that (C) and (D) are the correct options. Can someone please tell me where I am going wrong? Thank you.


Offline Borek

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Re: Stoichiometry question
« Reply #1 on: January 01, 2007, 06:24:23 AM »
How does the oxidation of FeC2O4 looks like?
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Offline Albert

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Re: Stoichiometry question
« Reply #2 on: January 01, 2007, 06:32:30 AM »
Moreover, why did you solve it using equivalents? It's easier to do it through traditional moles, isn't it?

Offline xstrae

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Re: Stoichiometry question
« Reply #3 on: January 01, 2007, 01:08:36 PM »
Fe2+ --- > Fe+3
so 'n' factor is 1 right? Am I doing anything else wrong? Because now, none of the options seem to match. thanks.

Quote
Moreover, why did you solve it using equivalents? It's easier to do it through traditional moles, isn't it?

I find equivalents much easier. Maybe its because I have had a lot of practice in that.

Offline Borek

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Re: Stoichiometry question
« Reply #4 on: January 01, 2007, 02:56:26 PM »
What about oxalate?
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Offline xstrae

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Re: Stoichiometry question
« Reply #5 on: January 02, 2007, 10:32:35 AM »
ohh sorry.

C23 ----> 2C+4  + 2e-

So 'n' factor is 3 right? Hence, (C) and (D). I understand now. Thanks a lot for your *delete me*

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