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Hydroxylamine...

Question

Hydroxylamine reduces iron (III) according to the equation,
$2 N H_{2} O H (a q) + 4 F e^{3 +} (a q) \to N_{2} O (g) ↑ + H_{2} O (l) + 4 F e^{2 +} (a q) + 4 H^{+} (a q)$
Iron (II) thus produced is estimated by titration with a standard permanganate solution. The reaction is:
${M n O_{4}}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
A $10 m L$ sample of hydroxylamine solution was diluted to one litre. $50 m L$ of this diluted solution was boiled with excess of iron (III) solution. The resulting solution required $12 m L$ of $0.02 M K M n O_{4}$ solution for complete oxidation of iron (II). Calculate the mass of hydroxylamine in one litre of the original solution.

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Solution

$12 m L$ of $0.02 M K M n O_{4} \equiv 12 m L$ of $0.1 N K M n O_{4}$
$\equiv 12 m L$ of $0.1 N F e^{2 +}$
$\equiv 12 m L$ of $0.1 N N H_{2} O H$
Eq. mass of $N H_{2} O H = \frac{M o l . m a s s}{2} = \frac{33}{2} = 16.5$
Mass of $N H_{2} O H$ in $12 m L$ of $0.1 N N H_{2} O H$ soln.
$= \frac{N \times E \times V}{1000} = \frac{0.1 \times 16.5 \times 12}{1000} = 0.0198 g$
$50 m L$ of diluted solution contains $N H_{2} O H = 0.0198 g$
$1000 m L$ of diluted solution contains $N H_{2} O H$
$= \frac{0, 0198}{50} \times 1000 = 0.396 g$
$10 m L$ of original solutions contains $N H_{2} O H = 0.396 g$
$1000 m L$ of original solution contains $N H_{2} O H = 100 \times 0.396 = 39.6 g$ .

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Q. Hydroxylamine reduces iron (III) according to the following reaction:

2 N H_{2} O H + F e^{3 +} \to N_{2} O (g) ↑ + H_{2} O + 4 F e^{2 +} + 4 H^{⨁}

Iron(II) thus produced is estimated by titration with a standard permanganate solution. The reaction is:

M n O_{4}^{⊝} + 5 F e^{2 +} + 8 H^{⨁} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

10

mL sample of hydroxylamine solution was diluted to

1

50

mL of this diluted solution was boiled with an excess of iron (III) solution, the resulting solution required

12

0.02 M

K M n O_{4}

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, M n = 55, F e = 56) :

Q. A sample of hydrazine sulphate

(N_{2} H_{6} S O_{4})

was dissolved in

100 m L

10 m L

of this solution was reacted with an excess of

F e C l_{3}

solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required

20 m L

\frac{M}{50} K M n O_{4}

. Estimate the amount of hydrazine sulphate in one litre of solution.
Reactions:

4 F e^{3 +} + N_{2} H_{4} \to N_{2} + 4 F e^{2 +} + 4 H^{+}

{M n O_{4}}^{-} + 5 F e^{2 +} + 8 H^{-} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

A sample of hydrazine sulphate

(N_{2} H_{6} S O_{4})

was dissolved in

100

10

mL of this solution was reacted with excess of

F e C l_{3}

solution and warmed to complete the reaction. Ferrous ions formed was estimated and it required

20

\frac{M}{50} K M n O_{4}

solution. Estimate the mass of hydrazine sulphate in one litre of solution.

4 F e^{3 +} + N_{2} H_{4} ⟶ N_{2} + 4 F e^{2 +} + 4 H^{+}

M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} ⟶ M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

A sample of hyhdrazine sulphate $(N_{2} H_{6} S O_{4})$ was dissolved in 100 ml water. 10 ml of this solution was reacted with excess of solution and warmed to complete the reaction. Ferrous ions formed required 20 ml of solution.
Given : $4 F e^{3 +} + N_{2} H_{4} \to N_{2} + F e^{2 +} + 4 H^{+}$
$M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$

The amount (g) of hydrazine sulphate in one litre is

Q. A sample of hydrazine sulphate

(N_{2} H_{6} S O_{4})

was dissolved in

100

mL of water and

10

mL of this solution was reacted with excess of ferric chloride solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required

20

\frac{M}{50}

potassium permanganate solution. Estimate the amount of hydrazine sulphate in

1

L of the solution. Reactions are given below:

4 F e^{3 +} + N_{2} H_{4} \to N_{2} + 4 F e^{2 +} + 4 H^{⨁}

M n O_{4}^{⊝} + 5 F e^{2 +} + 8 H^{⨁} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

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