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Faraday's Second Law

One mole of ...

Question

One mole of $K M n O_{4}$ is used for complete oxidation of $F e S O_{4}$ , $F e C_{2} O_{4}$ and $H_{2} C_{2} O_{4}$ respectively and separately. Pick up the correct statement.

A

5

moles of

F e S O_{4}

can be oxidized

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B

\frac{3}{5}

moles of

F e C_{2} O_{4}

can be oxidized

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C

\frac{5}{3}

moles of

F e C_{2} O_{4}

can be oxidized

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D

2.5

moles of

H_{2} C_{2} O_{4}

can be oxidized

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Solution

The correct options are
B $\frac{5}{3}$ moles of $F e C_{2} O_{4}$ can be oxidized
C $2.5$ moles of $H_{2} C_{2} O_{4}$ can be oxidized
D $5$ moles of $F e S O_{4}$ can be oxidized
The balanced reaction is as follows:
$M n O_{4}^{-} + 8 H^{+} + 5 F e^{+ 2} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
$5$ moles of $F e S O_{4}$ can be oxidized by one mole of $K M n O_{4}$ .
$2 K M n O_{4} + 5 H_{2} C_{2} O_{4} + 3 H_{2} S O_{4} \to 2 M n S O_{4} + 10 C O_{2} + 8 H_{2} O$
$2.5$ moles of $H_{2} C_{2} O_{4}$ can be oxidized by one mole of $K M n O_{4}$ .
$10 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 3 K_{2} S O_{4} + 6 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 24 H_{2} O + 20 C O_{2}$ .
Hence, $\frac{5}{3}$ moles of $F e C_{2} O_{4}$ can be oxidized by one mole.

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