Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified $K M n O_{4}$ for complete oxidation?

F e C_{2} O_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F e (N O_{2})_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F e S O_{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

F e S O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $F e S O_{4}$
$5 F e C_{2} O_{4} + 3 K M n O_{4} A c i d i c ⟶ 5 F e^{3 +} + 10 C O_{2} + 3 M n^{2 +}$

$5$ $m o l e s$ $F e C_{2} O_{4}$ requires $3$ $m o l e s$ $K M n O_{4}$

$∴ 1$ $m o l e$ $F e C_{2} O_{4}$ willl require $\frac{3}{5}$ $m o l e s$ $K M n O_{4}$

$5 F e (N O_{2})_{2} + 5 K M n O_{4} A c i d i c ⟶ 5 F e^{3 +} + 10 N O_{3}^{-} + 5 M n^{2 +}$

$5$ $m o l e s$ $F e (N O_{2})_{2}$ requires $5$ $m o l e s$ $K M n O_{4}$

$∴ 1$ $m o l e$ $F e (N O_{2})_{2}$ willl require $1$ $m o l e s$ $K M n O_{4}$

$5 F e S O_{4} + 1 K M n O_{4} A c i d i c ⟶ 5 F e^{3 +} + M n^{2 +}$

$5$ $m o l e s$ $F e S O_{4}$ requires $1$ $m o l e s$ $K M n O_{4}$

$∴ 1$ $m o l e$ $F e S O_{4}$ willl require $\frac{1}{5}$ $m o l e s$ $K M n O_{4}$

$5 F e S O_{3} + 3 K M n O_{4} A c i d i c ⟶ 5 F e^{3 +} + 5 S O_{4}^{2 -} + 3 M n^{2 +}$

$5$ $m o l e s$ $F e S O_{3}$ requires $3$ $m o l e s$ $K M n O_{4}$

$∴ 1$ $m o l e$ $F e S O_{3}$ willl require $\frac{3}{5}$ $m o l e s$ $K M n O_{4}$

$∴ F e S O_{4}$ will required least amount of $K M n O_{4}$ per mole $(\frac{1}{5})$

Hence, option C is correct.

Suggest Corrections