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In the redox reaction
$x {KMnO}_{4} + {NH}_{3} ⟶ y K {NO}_{3} + {MnO}_{2} + KOH + H_{2} O$

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$x = 4, y = 6$

$x = 3, y = 8$

$x = 8, y = 6$

$x = 8, y = 3$

answer is D.

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Detailed Solution

$8 {KMnO}_{4} + 3 {NH}_{3} ⟶ 8 {MnO}_{2} + 3 {KNO}_{3} + 5 KOH + 2 H_{2} O$

Note: This can be done by hit and trail method, i.e. substitute all the options and choose the correct one.

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