CameraIcon

SearchIcon

MyQuestionIcon

1

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

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

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

B

\frac{3}{5}

moles of

F e C_{2} O_{4}

can be oxidized

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

C

\frac{5}{3}

moles of

F e C_{2} O_{4}

can be oxidized

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

D

2.5

moles of

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

can be oxidized

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

Open in App

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.

Suggest Corrections

thumbs-up

0

Similar questions

Q. Number of moles of acidic

K M n O_{4}

needed for complete oxidation of

1

mole equimolar mixture of

F e S O_{4}

F e C_{2} O_{4}

F e S O_{4}

F e C_{2} O_{4}

is dissolved in

H_{2} S O_{4}

. The complete oxidation of sample required 8/3 eqv. of

K M n O_{4}

. After oxidation, the reaction mixture was reduced by

Z n

. On further oxidation by

K M n O_{4}

5/3 eqv were used. The moles of

F e C_{2} O_{4}

in the mixture is

3

F e S O_{4}

are oxidized by

a

K M n O_{4}

in acid medium where as

3

F e C_{2} O_{4}

are oxidized by

b

K M n O_{4}

in acid medium. The ratio of

a

b

Q. Two moles of

F e S O_{4}

are oxidized by

x

K M n O_{4}

in acid medium, whereas

3

F e C_{2} O_{4}

are oxidized by

y

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

in acid medium. The value of

\frac{x}{y}

Q. 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}

F e_{2} (S O_{4})_{3}

in acidic medium, the number of moles of

K M n O_{4}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Faraday's Laws

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Faraday's Second Law

Standard VII Chemistry

Join BYJU'S Learning Program

CrossIcon