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Balancing Redox Equations

A mixture of ...

Question

A mixture of $F e O$ and $F e_{2} O_{3}$ is completely reacted with $100 mL$ of $0.25 M$ acidified $K M n O 4$ solution. The resultant solution was then titrated with $Z n$ dust which converted the $F e^{3 +}$ ions to $F e^{2 +}$ ions. The $F e^{2 +}$ ions required $1000 mL$ of $0.10 M$ $K_{2} C r_{2} O_{7}$ solution. Find out the weight percentage of $F e_{2} O_{3}$ in the mixture.

80.85

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89.41

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Solution

The correct option is A 80.85
Milli-equivalents of $F e O$ is equal to
Milli-equivalents of $K M n O_{4}$
$= 0.25 \times 5 \times 100$

Milli-moles of $F e O (n = 1) = 0.25 \times 100 \times 5 = 125$

Total m-eq or m-mol of $F e^{2 +}$
$= 1000 \times 0.1 \times 6 = 600$

(from $F e O$ and $F e_{2} O_{3}$ after reaction with $Z n$ dust)

Milli-moles of $F e^{2 +}$ from $F e_{2} O_{3}$
$= 600 - 125 = 475$
Milli-moles of $F e_{2} O_{3} = \frac{475}{2}$
Weight of $F e O = 0.125 \times (56 - 16) g = 9 g$
Weight of $F e_{2} O_{3}$ $= \frac{475}{2} \times \frac{160}{1000} = 38 g$
$%$ of $F e_{2} O_{3} = \frac{38}{9 + 38} \times 100 = 80.85$

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Similar questions

Q. A mixture of

F e O

and

F e_{2} O_{3}

is completely reacted with

100 m L

0.25 M

acidified

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 +}

.
The

F e^{2 +}

required

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.

A quantity of

25.0

mL of solution containing both

F e^{2 +}

and

F e^{3 +}

ions is titrated with

25.0

mL of

0.02 M

K M n O_{4}

(in dilute

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

ions. Next

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

mL of the same

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 :

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.

A quantity of

25.0

mL of solution containing both

F e^{2 +}

and

F e^{3 +}

ions is titrated with

25.0

mL of

0.02 M

K M n O_{4}

(in dilute

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

ions. Next

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

mL of the same

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

Above titration takes place in acidic medium. Acid chosen is :

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.

A quantity of

25.0

mL of solution containing both

F e^{2 +}

and

F e^{3 +}

ions is titrated with

25.0

mL of

0.02 M

K M n O_{4}

(in dilute

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

ions. Next

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

mL of the same

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

Zinc added in the second titration will :

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.

A quantity of

25.0

mL of solution containing both

F e^{2 +}

and

F e^{3 +}

ions is titrated with

25.0

mL of

0.02 M

K M n O_{4}

(in dilute

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

ions. Next

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

mL of the same

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

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

is used instead of

0.02 M K M n O_{4}

, its volume required in these titrations will be respectively :

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