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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iodine is oxidised

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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
B $H_{2} O$ is one of the products
C iodine is oxidised
D stoichiometric coefficient of ${H S O_{4}}^{-}$ is $6$
The balanced equation is $C l O_{3}^{-} + 6 I^{-} + 6 H_{2} S O_{4} \to 3 I_{2} + C l^{-} + 6 H S O_{4}^{-} + 3 H_{2} O$ .
Here, iodine is oxidised and stoichiometric coefficient of $H S O_{4}^{-}$ is $6$ . Moreover, water is present on product side.

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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}

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}

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