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The values of x, y and z in the reaction are respectively :
$4 {KO}_{2} + x H_{2} O + y {CO}_{2} ⟶ 4 {KHCO}_{3} + z O_{2}$

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An Intiative by Sri Chaitanya

3, 2, 5

3, 6, 6

2, 4, 3

4, 3, 6

answer is A.

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Detailed Solution

$\begin{array}{l} O = 8 + x + 2 y = 12 + 22 \\ x + 2 y = 4 + z \dots (i) \\ H = 2 x = 4 \\ X = 2 - (ii) \\ c = y = 4 - (iii) \\ 2 + 2 \times 4 = 4 + 12 \\ 10 = 4 + 12 \\ 2 = 3 \end{array}$

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