The number of moles of $M n O_{4}^{-}$ required to oxidize one mole of ferrous oxalate completely in an acidic medium will be:

7.5

mole

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

0.2

mole

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

0.6

mole

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

0.4

mole

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

Open in App

Solution

The correct option is C $0.6$ mole
$10 F e C_{2} O_{4} + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 3 K_{2} S O_{4} + 6 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 24 H_{2} O + 20 C O_{2}$

6 moles of $M n O_{4}^{-}$ are required to oxidise 10 moles of ferrous oxalate, then the number of moles of $M n O_{4}^{-}$ required to oxidize one mole of ferrous oxalate completely in an acidic medium = $\frac{1 \times 6}{10} = 0.6$ moles of $M n O_{4}^{-}$

Suggest Corrections

Similar questions

M n O_{4}^{-}

1 mole of equimolar mixture of ferric oxalate and ferrous oxalate will require x mole of $K M n O_{4}$ in acidic medium for complete oxidation, x is

K M n O_{4}

K M n O_{4}

Join BYJU'S Learning Program