For the reaction:I^{-} + ClO_{3}^{-} + H_{2}S | vedclass

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(B) The balanced chemical equation is:$6I^{-} + ClO_{3}^{-} + 6H_{2}SO_{4} \longrightarrow Cl^{-} + 6HSO_{4}^{-} + 3I_{2} + 3H_{2}O$$1$. The stoichiom

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$A, B, D$

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(B) The balanced chemical equation is:$6I^{-} + ClO_{3}^{-} + 6H_{2}SO_{4} \longrightarrow Cl^{-} + 6HSO_{4}^{-} + 3I_{2} + 3H_{2}O$$1$. The stoichiometric coefficient of $HSO_{4}^{-}$ is $6$. (Statement $A$ is correct)$2$. The oxidation state of $I$ changes from $-1$ to $0$,so $I^{-}$ is oxidized. (Statement $B$ is correct)$3$. The oxidation state of $S$ in $H_{2}SO_{4}$ is $+6$ and in $HSO_{4}^{-}$ is $+6$. Thus,sulphur is not reduced. (Statement $C$ is incorrect)$4$. $H_{2}O$ is formed as a product. (Statement $D$ is correct)Therefore,the correct statements are $A, B, D$.

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For the reaction:$I^{-} + ClO_{3}^{-} + H_{2}SO_{4} \longrightarrow Cl^{-} + HSO_{4}^{-} + I_{2}$The correct statement$(s)$ in the balanced equation is/are:$A$. Stoichiometric coefficient of $HSO_{4}^{-}$ is $6$.$B$. Iodide is oxidized.$C$. Sulphur is reduced.$D$. $H_{2}O$ is one of the products.