NaIO3 reacts with NaHSO3 according to following equation IO3-3HSO3- - I-3H-3SO2-4The equivalent weight of oxidant molar mass - M in the reaction is-

The correct option is D M6 IO^ _3+3HSO^ _3 ⟶ I^ +3H^++3SO_4^2 Oxidation state of I in IO^ _3 ⇒ x 6= 1 #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Chemistry

Potassium Permanganate

NaIO3 reacts ...

Question

$N a I O_{3}$ reacts with $N a H S O_{3}$ according to following equation
$I O_{3}^{-} + 3 H S O_{3}^{-} \to I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}$
The equivalent weight of oxidant (molar mass = $M$ ) in the reaction is:

\frac{M}{2}

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\frac{M}{3}

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\frac{M}{1}

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\frac{M}{6}

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Solution

The correct option is D $\frac{M}{6}$
$I O_{3}^{-} + 3 H S O_{3}^{-} ⟶ I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}$
Oxidation state of $I$ in $I O_{3}^{-} \Rightarrow x - 6 = - 1$
$x = + 5$
Oxidation state of $I$ in $I^{-} \Rightarrow - 1$
Thus $I$ is getting reduced from $+ 5$ to $- 1$
Therefore $N a I O_{3}$ is an oxidant or oxidising agent. Change is oxidation state per $I$ is $+ 6$
$∴$ $1$ mole of $I$ is absorbing $6 e^{-}$
Equivalent mass= $\frac{Molar mass}{n_{f}} = \frac{M}{6}$

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Similar questions

N a I O_{3}

reacts with

N a H S O_{3}

according to equation

I O_{3}^{-} + 3 H S O_{3}^{-} \to I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}

The weight of

N a H S O_{3}

required to react with

100 m L

of solution containing

0.68 g

N a I O_{3}

I O_{3}^{-} - --- - + 3 H S O_{3}^{-} - ------- - ⟶ I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}

The equivalent weight of the underlined species by taking its molecular weight as

M

is :

N a I O_{3}

reacts with

N a H S O_{3}

according to equation

I O_{3}^{-} + 3 H S O_{3}^{-} \to I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}

The weight of

N a H S O_{3}

required to react with

100 m L

of solution containing

0.68 g

N a I O_{3}

Chile salt petre is a natural source of

N a N O_{3}

which also contains

N a I O_{3}

. The

N a I O_{3}

can be used as a source of iodine which produced in the following reactions:

I O_{3}^{-} + 3 H S O_{3}^{-} ⟶ I^{-} + 3 H^{+} 3 S O_{4}^{2 -}

... (i)

5 I^{-} + I O_{3}^{-} + 6 H^{+} ⟶ 3 I_{2} (s) + 3 H_{2} O

... (ii)

1.0

L of the starting solution which contains

5.80

N a I O_{3}

per litre is treated with stoichiometric quantity of

N a H S O_{3}

. A further quantity of the starting solution is added to the reaction mixture to bring about the second reaction.

(Atomic mass of

I = 127, N a = 23, S = 32

)

___________ equivalents of

I O_{3}^{-}

are present in one mole of

N a I O_{3}

in step (i).

Chile salt petre is a natural source of

N a N O_{3}

which also contains

N a I O_{3}

. The

N a I O_{3}

can be used as a source of iodine produced in the following reactions:

I O_{3}^{-} + 3 H S O_{3}^{-} ⟶ I^{-} + 3 H^{+} 3 S O_{4}^{2 -}

... (i)

5 I^{-} + I O_{3}^{-} + 6 H^{+} ⟶ 3 I_{2} (s) + 3 H_{2} O

... (ii)

1.00

L of the starting solution which contains

5.80

N a I O_{3}

per litre is treated with stoichiometric quantity of

N a H S O_{3}

. Further, a quantity of the starting solution is added to the reaction mixture to bring about the second reaction.

[I = 127, N a = 23, S = 32]

How many equivalents of

I O_{3}^{-}

are used in step (ii)?

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NaIO3 reacts with NaHSO3 according to following equationIO−3+3HSO−3→I−+3H++3SO2−4The equivalent weight of oxidant (molar mass = M) in the reaction is:

$N a I O_{3}$ reacts with $N a H S O_{3}$ according to following equation
$I O_{3}^{-} + 3 H S O_{3}^{-} \to I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}$
The equivalent weight of oxidant (molar mass = $M$ ) in the reaction is: