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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.

A

80.85

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B

19.15

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C

50

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D

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

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.

25

mL of a solution of

F e^{2 +}

ions was titrated with a solution of the oxidising agent

C r_{2} O_{7}^{2 -}

32.45

0.0153

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

solution was required. What is the molarity of the

F e^{2 +}

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 +}

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{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 :

25

mL of a solution of

F e^{2 +}

ions was titrated with a solution of the oxidising agent

C r_{2} O_{7}^{2 -} . 32.45

0.0153

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

solutin was required. What is the molarity of the

F e^{2 +}

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

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

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