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Ideal Gas Equation

1 mol of Fe...

Question

$1$ mol of $F e^{2 +}$ is oxidized to $F e^{3 +}$ using $K_{2} C r_{2} O_{7}$ and $K M n O_{4}$ solutions separately. Which of the following are correct for the titration of $F e^{2 +}$ to $F e^{3 +}$ ?

A

V_{K_{2} C r_{2} O_{7}} : V_{K M n O_{4}} : : 5 : 6

, if both have molarity

1 M

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B

V_{K_{2} C r_{2} O_{7}} : V_{K M n O_{4}} : : 6 : 5

, if both have molarity

1 M

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C

Moles of

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

Moles of

K M n O_{4}

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D

Equivalents of

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

Equivalents of

K M n O_{4}

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Solution

The correct options are
A $V_{K_{2} C r_{2} O_{7}} : V_{K M n O_{4}} : : 5 : 6$ , if both have molarity $1 M$
B Equivalents of $K_{2} C r_{2} O_{7} =$ Equivalents of $K M n O_{4}$
The involved reactions are given below:
$2 K_{2} C r_{2} O_{7} + 3 F e S O_{4} + 14 H_{2} S O_{4} \to 2 K_{2} S O_{4} + 2 C r_{2} (S O_{4})_{3} + 3 F e (S O_{4})_{3} + 14 H_{2} O$
$M n O_{4}^{-} + 8 H^{+} + 5 F e^{+ 2} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
As we know, for every reaction, the number of equivalents of reactants and products are the same.
So, equivalents of $K_{2} C r_{2} O_{7} =$ equivalents of $K M n O_{4}$
Also, from balanced reactions, it can be seen that,
$V_{K_{2} C r_{2} O_{7}} : V_{K M n O_{4}} : : 5 : 6$ , if both have molarity $1 M$ .

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Similar questions

Q. If equal volumes of

1 M

K M n O_{4}

1 M

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

solutions are used to oxidise

F e^{2 +}

in acidic medium, then

F e^{2 +}

will be oxidised

Q. If two sample of 2 mole ferrous oxalate is treated separately with
(i) Acidic KMnO4 and moles of KMnO4 used are x
(ii) Acidic K2Cr2O7 and moles of K2Cr2O7 used are y

Q. Equal volumes of 1 M each of

K M n O_{4}

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

are used to oxidise Fe(II) solution in acidic medium. The amount of Fe oxidized will be:

Q. A mixture of

F e O

F e_{2} O_{3}

is completely reacted with

100 m L

0.25 M

K M n O_{4}

solution. The resultant solution was then titrated with

Z n

dust which converted

F e^{3 +}

of the solution to

F e^{2 +}

F e^{2 +}

1000 m L

0.10 M

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

solution. Find out the weight

% F e_{2} O_{3}

in the mixture.

Q. The following titration method is given to determine the total content of the species with variable oxidation states. Answer the question given at the end of it.

25.0

mL of solution containing both

F e^{2 +}

F e^{3 +}

ions is titrated with

25.0

0.02 M

K M n O_{4}

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

K M n O_{4}

solution for oxidation to

F e^{3 +}

.

{M n O_{4}}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

The molar concentration of

F e^{2 +}

in the original solution is :

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