Milliequivalents of $F e C_{2} O_{4}$ reacted is:

6

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20

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40

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none of the above

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Solution

The correct option is B $20$
20 mL of $\frac{M}{10}$ $B a (M n O_{4})_{2}$ corresponds to $\frac{M}{10} \times 20 m L = 2$ mmoles.
In acidic medium, one $M n O_{4}^{-}$ ion takes 5 electrons. One $B a (M n O_{4})_{2}$ molecule will take $2 \times 5 = 10$ electrons. Hence, 2 mmoles of $B a (M n O_{4})_{2}$ corresponds to $2 \times 10 = 20$ milliequivalents.
20 milliequivalents of $B a (M n O_{4})_{2}$ will react with 20 milliequivalents of $F e C_{2} O_{4}$ .

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Similar questions

F e C_{2} O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + C O_{2} + M n S O_{4} + K_{2} S O_{4} + H_{2} O

n -

F e C_{2} O_{4}

F e C_{2} O_{4}

K M n O_{4}

¯ ¯¯¯¯¯ ¯ R S

¯ ¯¯¯¯¯¯ ¯ T U

S e O_{3}^{2 -} =

B r O_{3}^{⊝} -

B r O_{3}^{⊝}

B r O_{3}^{⊝} =

A s O_{2}^{⊝}

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