Hydrazine reacts with KIO3 in presence of HCl as- N2H4-IO-3-2H-Cl- ICl-N2-3H2O-The equivalent masses of N2H4 and KIO3 respectively are-

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Standard XII

Chemistry

Preparations and Reactions of Sodium Oxide

Hydrazine rea...

Question

Hydrazine reacts with $K I O_{3}$ in presence of HCl as:
$N_{2} H_{4} + I O_{3}^{-} + 2 H^{+} + C l^{-} \to I C l + N_{2} + 3 H_{2} O$ .
The equivalent masses of $N_{2} H_{4}$ and $K I O_{3}$ respectively are:

8, 87

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8, 35.6

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16, 53.5

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8, 53.5

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Solution

The correct option is A $8, 53.5$
Oxidation and reduction half reactions are:
$N_{2} H_{4} \to N_{2} + {4 H}^{+} + {4 e}^{-} {I O_{3}}^{-} + {6 H}^{+} + {5 e}^{-} \to I C l + 3 H_{2} O$
Equivalent mass is molecular mass divided by number of electrons gained or lost.
Equivalent mass of $N_{2} H_{4}$ is $\frac{32}{4} = 8$

Equivalent mass of $K I O_{3}$ is $\frac{214}{4} = 53.5$

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Similar questions

Q. Hydro-zine reacts with

K I O_{3}

in presence of

H C l

N_{2} H_{4} + I O_{3}^{-} + 2 H^{+} + C l^{-} \to I C l + N_{2} + 3 H_{2} O

The equivalent masses of

N_{2} H_{4}

and

K I O_{3}

N_{2} H_{4}

reduces

I O_{3}^{-}

I^{+}

in acidic medium as :

N_{2} H_{4} + I O_{3}^{-} + H^{+} ⟶ N_{2} + I^{+} + H_{2} O

Equivalent mass of

K I O_{3}

(molar mass = 214 g mol

^{- 1}

) is :

Q. A

50 m L

1.92 %

w / v

solution of a metal ion

M^{n +} a t . w t = 60

was treated with

5.32 g

hydrazinehydrate

(N_{2} H_{4} . H_{2} O) 90 % p u r e

and the mixture was saturated with

C O_{2}

gas when entire metal gets pecipitated as a complex

[M (N H_{2} - N H C O O)_{n}]

the complex was filtered of and the filterate was titrated with

M / 10 K I O_{3}

in the presence of

c o n c . H C l

according to the following equation :

N_{2} H_{4} + I O_{3}^{-} + 2 H^{+} + C l^{-} \to I C l + 3 H_{2} O + N_{2} ↑

The volume of

M / 10 K I O_{3}

solution needed for end point to arrive was

480 m L

. Find the value of

n

Q. In the following reaction, hydrazine is oxidised to

N_{2}

N_{2} H_{4} + O H^{-} \to N_{2} + H_{2} O + e^{-}

.
The equivalent mass of

N_{2} H_{4}

(hydrazine) is:

N_{2} H_{4}

reduces

I O_{3}^{-}

I^{+}

in acidic medium as

N_{2} H_{4} + I O_{3}^{-} + H^{+} ⟶ N_{2} + I^{+} + H_{2} O

N_{2} H_{4}

is oxidised to

N_{2}

. Both

N -

atom in

N_{2}

are from

N_{2} H_{4}

. It can be verified by :

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Hydrazine reacts with KIO3 in presence of HCl as:N2H4+IO−3+2H++Cl−→ICl+N2+3H2O.The equivalent masses of N2H4 and KIO3 respectively are:

The correct option is A 8,53.5Oxidation and reduction half reactions are:N2H4→N2+4H++4e−IO3−+6H++5e−→ICl+3H2OEquivalent mass is molecular mass divided by number of electrons gained or lost.Equivalent mass of N2H4 is 324=8Equivalent mass of KIO3 is 2144=53.5

Hydrazine reacts with $K I O_{3}$ in presence of HCl as:
$N_{2} H_{4} + I O_{3}^{-} + 2 H^{+} + C l^{-} \to I C l + N_{2} + 3 H_{2} O$ .
The equivalent masses of $N_{2} H_{4}$ and $K I O_{3}$ respectively are: