Question

$KO _{2}$, 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?

• A. $\ce{ KOH, H_2O_2}$
• B. $K _{2} O _{2}, H _{2} O _{2}$
• C. $\ce{ KOH, O_2}$
• D. $\ce{ H_2O_2 , O_2}$

Answer 1

Answer: (D)

$\ce{ H_2O_2 , O_2}$

Answer 2

(i) When $KO _{2}$ reacts with water, it gives $(A), (B)$ and $(C)$ as follows: $\underset{(B)}{2 KO _{2}}+2 H _{2} O \longrightarrow \underset{(A)}{2 KOH }+\underset{(B)}{ H _{2} O _{2}}+ O _{2}$ (ii) When $(B)$, i.e. $H _{2} O _{2}$ reacts with iodine in basic medium, it gives $(C)$, i.e. $O _{2}$, as shown below : $\underset{(B)}{ H _{2} O _{2}}+ I _{2}+2 KOH \longrightarrow 2 KI +2 H _{2} O + \underset{(C)}{O _{2}}$ $\because KOH$ can not reduce $I _{2}$,