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Hydroxylamine reduces iron (III) according to the equation: 2NH 2 OH + 4 Fe 3+ →N 2 O(g) ↑ + H 2 O + 4 Fe 2+ + 4H + Iron (II) thus produced is estimated by titration with standard permanganate solution. The is : MnO - 4 + 5 Fe 2+ + 8H + → Mn 2+ + 5 Fe 3+ + 4H 2 O A 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 KMnO 4 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)

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+ + 4H2O
A 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)               

Grade:10

1 Answers

Jitender Pal
askIITians Faculty 365 Points
9 years ago
(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

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