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Balancing Redox Equations

x mmol of KMn...

Question

x mmol of $K M n O_{4}$ reacts completely with y mmol of $M n S O_{4}$ in the presence of fluoride ions to give $M n F_{4}$ quantitatively, then:

x = 4

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4x = 1

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x > y

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x < y

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Solution

The correct option is D x < y
$K M n O_{4} n = 3 + M n S O_{4} n = 2 F^{-} - - \to M n F_{4}$

Meq. of $K M n O_{4}$ = Meq. of $M n S O_{4}$
3x = 2y
Hence, x < y

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