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Byju's Answer

Standard XI

Chemistry

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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Similar questions

Q. 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:

Q. 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:

500 ml of 0.1 M KCl, 200 ml of 0.01 M $N a N O_{3}$ and 500 ml of 0.1 M $A g N O_{3}$ was mixed. What is the total number of moles of ions in solution?

x

mmol of

{X e F}_{4}

quantitatively oxidized KI to

I_{2}

and liberated

X e

, along with formation of

K F

. This iodine required 20 ml of decinormal hypo solution for exact titration. The value of

x

is:

A_{2} O_{n}

is oxidized to

A O_{3}^{-}

K M n O_{4}

in acidic medium. If

1.34

mmol of

A_{2} O_{n}

requires

32.2

ml of

0.05

M acidified

K M n O_{4}

solution for complete oxidation. Find the value of

n

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x mmol of KMnO4 reacts completely with y mmol of MnSO4 in the presence of fluoride ions to give MnF4 quantitatively, then:

The correct option is D x < yKMnO4n=3+MnSO4n=2F−−−→MnF4 Meq. of KMnO4 = Meq. of MnSO4 3x = 2y Hence, x < y

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:

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