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Introduction to Oxidation and Reduction

0.1 M solutio...

Question

0.1 M solution of $K I$ reacts with excess of $H_{2} S O_{4}$ and $K I O_{3}$ solutions, according to the equation $5 I^{-} + I O_{3}^{-} + 6 H^{+} \to 3 I_{2} + 3 H_{2} O$
which of the following statements is/are correct

200 ml of the

K I

solution reacts with 0.004 mole

K I O_{3}

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100 ml of the

K I

solution reacts with 0.006 mole

H_{2} S O_{4}

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0.5 litre of the

K I

solution produced 0.005 mole of

I_{2}

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Equivalent weight of

K I O_{3}

is equal to

(\frac{m o l e c u l a r w e i g h t}{5})

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Solution

The correct options are
A 200 ml of the $K I$ solution reacts with 0.004 mole $K I O_{3}$
D Equivalent weight of $K I O_{3}$ is equal to $(\frac{m o l e c u l a r w e i g h t}{5})$
a) $200 m l$ of the $K I$ solution = $200 \times 0.1 m m o l o f K I = 20 m m o l$ .
So from the stoichiometry of the reaction, 200 ml of the $K I$ solution reacts with $4 m m o l$ of $K I O_{3}$ .
b) $100 m l$ of the $K I$ solution = $100 \times 0.1 m m o l o f K I = 10 m m o l$ .
So from the stoichiometry of the reaction, 100 ml of the $K I$ solution reacts with $\frac{6}{5} \times 10 = 12 m m o l o f H_{2} S O_{4}$ .
c) $0.5 l i t r e$ of the $K I$ solution = $500 \times 0.1 m m o l o f K I = 50 m m o l$ .
So from the stoichiometry of the reaction, 0.5 litre of the $K I$ solution produced $\frac{3}{5} \times 50 = 30 m m o l o f I_{2}$ .
d) Equivalent weight of $K I O_{3}$ is equal to $(\frac{m o l e c u l a r w e i g h t o f K I O_{3}}{n - f a c t o r o f K I O_{3}}) = \frac{m o l e c u l a r w e i g h t}{5}$

Suggest Corrections

Similar questions

0.1 M

H_{2} S O_{4}

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5 I^{-} + I O_{3}^{-} + 6 H^{+} \to 3 I_{2} + 3 H_{2} O

0.1 M

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{N a}_{2} S_{2} O_{3}

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