CameraIcon

SearchIcon

MyQuestionIcon

1

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

Hydrogen Peroxide

Oxalic acid, ...

Question

Oxalic acid, $H_{2} C_{2} O_{4}$ , reacts with paramagnet ion according to the balanced equation $5 H_{2} C_{2} O_{4} (a q) + 2 M n O_{4}^{-} (a q) ⇌ 2 M n^{2 +} (a q) + 10 C O_{2} (g) + 8 H_{2} O (l)$ . The volume in mL of $0.0162$ M $K M n O_{4}$ solution required to react with $25.0$ mL of $0.022$ M $H_{2} C_{2} O_{4}$ solution is:

A

13.6

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

B

18.5

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

C

33.8

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

D

84.4

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

Open in App

Solution

The correct option is A $13.6$
In the reaction
$+ 7 M n O_{4}^{-} ⟶ {M n}^{+ 2}$
thus $n f = 5$
${+ 3 C}_{2} O_{4}^{2 -} ⟶ {+ 4 C O}_{2}$
$2. C^{+ 3} ⟶ C^{+ 4}$
$n f = 2 \times 1 = 2$
thus, by equation
$({K M n O}_{4}) M_{1} V_{1} {n f}_{1} = M_{2} V_{2} {n f}_{2} (H_{2} C_{2} O_{4})$
$0.0162 M \times V_{1} \times 5 = 0.022 M \times 25.0 m L \times 2$
$V_{1} = 13.58 m L ≃ 13.6 m L$
Thus $13.58$ $m L$ of ${K M n O}_{4}$ solution is required to react with $25.0$ $m L$ of $0.022 M$ $H_{2} C_{2} O_{4}$ solution.

Suggest Corrections

thumbs-up

0

Similar questions

K M n O_{4}

reacts with oxalic acid according to the equation:

2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

K M n O_{4}

will react with?

1.0

K M n O_{4}

solution reacts with

0.125

F e^{2 +}

1.0

K H C_{2} O_{4} . H_{2} C_{2} O_{4}

solution reacts with

0.175

K M n O_{4}

solution, 8.4 ml volume of

0.2 M N a O H

which will be used to neutralise

1.0

mL of tetroxalate solution (all the three protons in tetroxalate are titrable).

K M n O_{4}

reacts with oxalic acid according to the equation

2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

20 m L

K M n O_{4}

is equivalent to:

20

0.2 M N a O H (a q)

solution is mixed with

35

0.1 M N a O H (a q)

solution and the resultant solution is diluted to

100

40

mL of this diluted solution reacted with

10 %

impure sample of oxalic acid

(H_{2} C_{2} O_{4})

. The weight of impure sample is :

(H_{2} C_{2} O_{4})

is present in many plants and vegetables. If

24.0

0.01

K M n O_{4}

solution is needed to titrate

1.0

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

to the equivalence-point, the mass percentage (as nearest integer) of

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

in the sample is :

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Hydrogen Peroxide

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Hydrogen Peroxide

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon