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`        Hydroxylamine reduces iron (III) according to the equation:2NH2OH + 4 Fe3+→N2O(g) ↑ + H2O + 4 Fe2+ + 4H+Iron (II) thus produced is estimated by titration with standard permanganate solution. The is :MnO-4 + 5 Fe2+ + 8H+→ Mn2+ + 5 Fe3+ + 4H2OA 10 ml. sample of hydroxylamine solution was dilute to 1 litre. 50 ml. of this diluted solution was boiled with an excess of iron (III) solution. The resulting solution required 12 ml. of 0.02 M KMnO4 solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in one litre of the original solution. (H = 1, N = 14, O = 16, K = 39, Mn = 55, Fe = 56)               `
5 years ago

Jitender Pal
365 Points
```
(i) Write balanced reaction for changes.

(ii) M. eq. of Fe2+ formed = M. eq. KMnO4ised

= V in ml * molarity * x

The given redox changes are

Fe3+ + e- → Fe2+

2N- → 2 N+ + 2e-

[In NH2OH, O.S. of N = - 1 and in N2O, O.S. of N = + 1. For two molecules of NH2OH, electron involved = 4]

Mn7+ + 5e- → Mn2+

= Meq. Of KMnO4 used = Vol. in ml * Molarity * x

= 12 * 0.02 * 5 =12 [50/10 = 5]

∴Meq.of Fe2+ formed by NHH2OH in 1000 ml of dil. Solution

= 1.2 * 1000/50 = 24

Meq.of original solution

= Meq. of NH2OH in 1000 ml. of dilute solution

WNH base 2OH/33/2 * 1000 = 24 [Eq. wt = 33/2]

WNH base 2OH = 24 *16.5/1000 = 0.396 g

∵ Wt. of NH2OH in 10 ml. of original solution = 0.396 g

∴ Wt. of NH2OH in 1 litre of original solution

= 0.396 * 100 = 39.6 g l-1

ALTERNATIVESOLTUION:

Given 2NH2OH + 4Fe3+ → N2O + H2­O + 4Fe2+ + 4H+ …..(i)

And MnO-4 + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O  ...(ii)

∴ 10NH2OH +4MnO-4 + 12H+ → 5N­2O + 21H2O + 4Mn2+

[On multiplying (i) by 5 and (ii) by 4 and then adding the resulting equations]

Molecular weight of NH2OH = 33

Thus 400 ml of 1M MnO4- would react with NH2OH = 330g

∴ 12 ml of 0.02 M KMnO4 would react with NH2OH = 330 * 12 *0.02/400 g

∴ Amount of NH2OH present in 1000 ml of dilute solution 330 *12 *0.02 *1000/400 *50 g

Since 10 ml of sample of hydroxylamine is dilute to one litre

∴ A mount of hydroxyl amine in one litre of original solution = 330 *0.02 *12 *1000/4000 *50 * 1000/10 g = 39*6 g

```
5 years ago
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