On boiling an aqueous solution of KClO_3 with | vedclass

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Question

(A) When an aqueous solution of $KClO_3$ is boiled with iodine $(I_2)$,iodine acts as a reducing agent and gets oxidized to iodate $(IO_3^-)$ while ch

Vedclass · Accepted Answer

$KIO_3$

Vedclass · Answer

(A) When an aqueous solution of $KClO_3$ is boiled with iodine $(I_2)$,iodine acts as a reducing agent and gets oxidized to iodate $(IO_3^-)$ while chlorate $(ClO_3^-)$ is reduced to chloride $(Cl^-)$.The balanced chemical equation for this reaction is:$2KClO_3 + I_2 	o 2KIO_3 + Cl_2$Therefore,the product obtained is $KIO_3$.

On boiling an aqueous solution of $KClO_3$ with iodine,the following product is obtained:

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