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Question

Hydroxylamine reduces iron (III) according to the following reaction:
2NH2OH+Fe3+N2O(g)+H2O+4Fe2++4H

Iron(II) thus produced is estimated by titration with a standard permanganate solution. The reaction is:
MnO4+5Fe2++8HMn2++5Fe3++4H2O
A 10 mL sample of hydroxylamine solution was diluted to 1 L. When 50 mL of this diluted solution was boiled with an excess of iron (III) solution, the resulting solution required 12 mL of 0.02M KMnO4 solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in 1 L of the original solution (H=1,N=14,O=16,K=39,Mn=55,Fe=56):

A
40gL1
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B
80gL1
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C
22gL1
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D
62gL1
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Solution

The correct option is B 40gL1
1molMnO45molFe2+2.5molNH2OH
12 mL of 0.02 M KMnO4 solution contains 12mL×0.02M=0.24mmol KMnO4.
It also corresponds to 0.24mmol×2.5=0.6mmol of NH2OH.
These many moles are present in 50 mL of diluted solution.
Hence, 1 L of diluted solution contains 100050×0.6mmol=12mmol=0.012mol
The molar mass of NH2OH is 14+3+16=33g/mol.
0.012mol of NH2OH corresponds to 0.012mol×33g/mol=0.396g.
This much NH2OH is present in the 10 mL of original solution.
Hence, 1 L of original solution contains 100010×0.396g=39.6g40g

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