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Stoichiometric Calculations

A 50 mL 1.92%...

Question

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 $n$ .

A

n = 1

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B

n = 2

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C

n = 3

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D

n = 4

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Solution

The correct option is C $n = 3$
$n_{M^{n +}} = \frac{1.92}{2 \times 60} m o l = 0.016 m o l$
$n h y d r a z i n e h y d r a t e = \frac{0.9 \times 5.332}{50} m o l = 0.096 m o l$
$n_{e q} K I O_{3} = 4 \times \frac{1}{10} N \times 0.48 L = 0.192 e q . = n_{e q .} hydrazine reacted with K I O_{3}$
$∴ n hydrazine left after reaction with M^{n +} = \frac{0.192}{4} M o l = 0.048 m o l$
$∴ n hydrazine hydrate reacted with M^{n +} = (0.096 - 0.048) m o l = 0.048 m o l$
$M^{n +} + n H_{2} N N H C O O^{-} \to [M (H_{2} N N H C O O)_{n}]$
$\frac{n_{M^{n +}}}{n_{h y d r a z i n e}} = \frac{1}{n} = \frac{0.016}{0.048} = \frac{1}{3}$
$∴ n = 3$

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Similar questions

Q. A 50 ml 1.92% (w/v) solution of a metal ion

M^{n +}

(atomic weight = 120) was treated with

6 g

hydrazinehydrate (80% pure) and mixture was saturated with

C O_{2}

gas when entire metal gets precipitated as a complex

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

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according to the following equation:

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

\frac{M}{10} K I O_{3}

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

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