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Standard XII

Chemistry

Stoichiometry

The product ...

Question

The product $(A)$ and $(B)$ are respectively:

K I, I_{2}

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I_{2}, K I O_{3}

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K I O_{3}, K I O_{4}

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I_{2}, I_{2}

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Solution

The correct option is A $I_{2}, K I O_{3}$
In Acidic medium:

$2 M n O_{4}^{-} + 10 I^{-} + 16 H^{+} ⟶ 2 M n^{+ 2} + 8 H_{2} O + 5 I_{2}$

In Basic medium:

$2 M n O_{4}^{-} + H_{2} O + I^{-} ⟶ 2 M n O_{2} + 2 ⊝ O H + I O_{3}^{-}$

Hence, the correct option is B.

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Similar questions

Q. Which two of the following salts are used for preparing iodized salt?
(i)

K I O_{3}

(ii)

K I

(iii)

I_{2}

(iv)

H I

Q. Consider the following statements for the reaction:

K I O_{3} + 5 K I + 6 H C l \to 3 I_{2} + 6 K C l + 3 H_{2} O

1.

I^{-}

is reduced to

I_{2}

2.

{I O_{3}}^{-}

is oxidized to

I_{2}

3.

{I O_{3}}^{-}

is reduced to

I_{2}

4.

Oxidation number of

I

increases from

- 1

(in

K I

) to zero (in

I_{2}

)

The correct statements are:

K I + I_{2} H N O_{3} \to H I O_{3} + K I O_{3} + N O_{2}

If 3 moles of KI and 2 moles of

I_{2}

are mixed with excess

H N O_{3}

, then volume of

N O_{2}

gas evolved at NTP is:

Q. If

x

moles of

K I

are oxidised by number of moles of

{K I O}_{3}

formed when 1 mole of

I_{2}

is boiled with excess of

K O H

, then what is the value of

\frac{1}{x}

6 K O H + 4 I_{2} ⟶ 5 K I + K I O_{3} + 3 H_{2} O

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

Q. Assertion :

K I O_{3}

reacts with

K I

to liberate iodine which is titrated with standard hypo solution. The reactions are:
(i)

I O_{3}^{-} \to I_{2}

(Valency factor = 5/3)
(ii)

I_{2} + S_{2} O_{3}^{2} \to S_{4} O_{6}^{2 -} + I^{-}

(Valence factor

=

2)
Milliequivalent of hypo

=

meq. of

I_{2} =

meq. of

I O_{3}^{-}

Milliequivalent of hypo =

2 \times m e q .

I O_{3}^{-} =

meq. of

I^{-}

Reason: Valency factor of

I_{2}

in both the equation are different, therefore we cannot compare milliequivalents in sequence.

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The product (A) and (B) are respectively:

The correct option is A I2,KIO3In Acidic medium:2MnO−4+10I−+16H+⟶2Mn+2+8H2O+5I2In Basic medium:2MnO−4+H2O+I−⟶2MnO2+2⊝OH+IO−3Hence, the correct option is B.

The product $(A)$ and $(B)$ are respectively:

The correct option is A $I_{2}, K I O_{3}$
In Acidic medium:

$2 M n O_{4}^{-} + 10 I^{-} + 16 H^{+} ⟶ 2 M n^{+ 2} + 8 H_{2} O + 5 I_{2}$

In Basic medium:

$2 M n O_{4}^{-} + H_{2} O + I^{-} ⟶ 2 M n O_{2} + 2 ⊝ O H + I O_{3}^{-}$

Hence, the correct option is B.