You visited us 1 times! Enjoying our articles? Unlock Full Access!

Types of Redox Reactions

The number of...

Question

The number of moles of $K M n O_{4}$ required to oxidise $1$ mol of $F e (C_{2} O_{4})$ is :

0.6

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

1.67

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

0.2

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

0.4

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

Open in App

Solution

The correct option is A 0.6
Equivalent of $M n O_{4}^{⊖} =$ Equivalent of $F e (C_{2} O_{4})$
$(n = 5) (n = 3)$
$∴ \frac{1}{5} m o l = \frac{1}{3} m o l$
or, $\frac{3}{5}$ mol of $M n O_{4}^{⊖} = 1$ mol of $F e (C_{2} O_{4})$
or, $0.6$ mol of $M n O_{4}^{⊖} = 1$ mol of $F e (C_{2} O_{4})$

Suggest Corrections

Similar questions

1

{C r}_{2} O_{7}^{2 -}

F e C_{2} O_{4}

1

F e C_{2} O_{4}

x

{C r}_{2} O_{7}^{2 -}

x

K M n O_{4}

K_{2} C r_{2} O_{7}

0.1

S n^{+ 4}

1

F e C_{2} O_{4}

0.6

M n O_{4}^{-}

M n O_{4}^{-}

{F e}^{2 +}

C_{2} O_{4}^{2 -}

2

F e C_{2} O_{4}

x

{C r}_{2} O_{7}^{2 -}

x

Join BYJU'S Learning Program