Q21 / 90 · JEE Main 2019 (08 Apr Shift 1) · Chemistry

In order to oxidize a mixture of one mole of each of $F e C_{2} O_{4}, F e_{2} {(C_{2} O_{4})}_{3}, F e S O_{4}$ and $F e_{2} {(S O_{4})}_{3}$ in acidic medium, the number of moles of $K M n O_{4}$ is:

A
$1$
Correct answer
B
$2$
C
$3$
D
$1.5$

Solution

E q . K M n O_{4} \Rightarrow E q . F e S O_{4} + E q . F e_{2} {(S O_{4})}_{3} + E q F e C_{2} O_{4} + E q . F e_{2} {(C_{2} O_{4})}_{3}

K M n O_{4} \Rightarrow M n O_{4}^{-}

5 e^{-} + 8 H^{+} + M n O_{4}^{-} \Rightarrow M n^{+ 2} + 4 H_{2} O

V.F = 5

So

E q . K M n O_{4} \Rightarrow E q . F e S O_{4} + E q . F e C_{2} O_{4} + E q . F e_{2} {(C_{2} O_{4})}_{3}

n \times 5 = 1 \times 1 + 1 \times 0 + 1 \times 3 + 1 \times 6

n \times 5 = 1 + 3 + 6

n = \frac{10}{5} = 2 m o l e

Hence, the correct answer is option: 'B'.

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