For the reaction:
$I^{-} + C l O_{3}^{-} + H_{2} S O_{4} \to C l^{-} + H S O_{4}^{-} + I_{2}$
The correct statement(s) in the balanced equation is/are :

stoichiometric coefficient of

H S O

_{4}^{-}

6

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iodide is oxidized

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sulphur is reduced

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H_{2} O

is one of the products

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Solution

The correct options are
A stoichiometric coefficient of $H S O$ $_{4}^{-}$ is $6$
B iodide is oxidized
D $H_{2} O$ is one of the products
The balanced equation is $C l O_{3}^{-} + 6 I^{-} + 6 H_{2} S O_{4} \to 3 I_{2} + C l^{-} + 6 H S O_{4}^{-} + 3 H_{2} O$ .
So, stoichiometric coefficient of $H S O_{4}^{-}$ is $6$ , iodide is oxidised from $I^{-}$ to $I_{2}$ and $H_{2} O$ is one of the product.

Options A,B and D are correct.

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Similar questions

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

I^{⊖} + C l O_{3}^{⊖} + H_{2} S O_{4} \to C l^{⊖} + H S O_{4}^{-} + I_{2}

I^{-} + C l O_{3}^{-} + H_{2} S O_{4} ⟶ C l^{-} + H S O_{4}^{-} + I_{2}

C l O^{-} ⇄ {C l O}_{3}^{-} + {C l}^{-}

{C l}^{-}

{C l O}_{3}^{-}

C l O^{-}

I^{⊖} + C l O_{3}^{⊖} + H_{2} S O_{4} \to C l^{⊖} + H S O_{4}^{-} + I_{2}

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