KO_2 reacts with water to form A,B and C. B f | vedclass

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(D) $(i)$ When $KO_2$ reacts with water,it gives $(A)$,$(B)$ and $(C)$ as follows:$2KO_2 + 2H_2O \rightarrow 2KOH + H_2O_2 + O_2$Here,$A = KOH$,$B = H

Vedclass · Accepted Answer

$H_2O_2, O_2$

Vedclass · Answer

(D) $(i)$ When $KO_2$ reacts with water,it gives $(A)$,$(B)$ and $(C)$ as follows:$2KO_2 + 2H_2O \rightarrow 2KOH + H_2O_2 + O_2$Here,$A = KOH$,$B = H_2O_2$,and $C = O_2$.$(ii)$ When $(B)$,i.e.,$H_2O_2$ reacts with iodine in basic medium,it gives $(C)$,i.e.,$O_2$,as shown below:$I_2 + H_2O_2 + 2OH^{-} \rightarrow 2I^{-} + 2H_2O + O_2$Thus,$B$ is $H_2O_2$ and $C$ is $O_2$. Therefore,option $(D)$ is the correct answer.

$KO_2$ reacts with water to form $A$,$B$ and $C$. $B$ forms $C$ when it reacts with iodine in basic medium. What are $B$ and $C$ respectively?

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