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Question

Which of the following liberates $$O_2$$ upon hydrolysis?

A
$$Na_2O_2$$
B
$$Li_2O_2$$
C
$$KO_2$$
D
$$Pb_3O_4$$
Solution
Verified by Toppr

Correct option is B. $$KO_2$$
(A) $$Pb_3O_4+H_2O\rightarrow $$ No reaction

(B) $$2KO_2+2H_2O\rightarrow 2KOH +H_2O_2+O_2$$

(C) $$Na_2O_2+2H_2O\rightarrow 2NaOH + H_2O_2$$

(D) $$Li_2O_2+2H_2O\rightarrow 2LiOH + H_2O_2$$
So, option B is the correct answer.

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