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Q.

In the redox reaction: xKMnO4 + yNH3 → KNO3 + MnO2 + KOH + H2O

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a

x = 4, y = 6

b

x = 3, y = 8

c

x = 8, y = 3

d

x = 8, y = 6

answer is D.

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Detailed Solution

\large \mathop {xKMn{O_4}}\limits^{ + 7} + {\text{ }}\mathop {yN{H_3}}\limits^{ - 3} \to {\text{ }}\mathop {KN{O_3}}\limits^{ + 5} + {\text{ }}\mathop {Mn{O_2}}\limits^{ + 4} + {\text{ }}KOH{\text{ }} + {\text{ }}{H_2}O
(K+ ions are the spectator ions and the medium is basic)
\large \mathop {MnO_4^ - }\limits_{ + 7} \; \to \;\mathop {Mn{O_2}}\limits_{ + 4}
Decrease in Ox.no = 3
\large \mathop {N{H_3}}\limits_{ - 3} \; \to \;\mathop {NO_3^ - }\limits_{ + 5}

decrease in ox.no=8
Criss cross method
8 MnO4- + 3NH34
x = 8 & y = 3

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