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Standard XII

Chemistry

Stoichiometric Calculations

A mixture of ...

Question

A mixture of $F e O$ and $F e_{2} O_{3}$ is completely reacted with $100 m L$ of $0.25 M$ 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 $1000 m L$ of $0.10 M$ $K_{2} C r_{2} O_{7}$ solution. Find out the weight $% F e_{2} O_{3}$ in the mixture.

80.85

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19.15

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50

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89.41

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Solution

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

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

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 $F e O = \frac{125 \times (56 + 16)}{1000} g = 9 g$

Weight of $F e_{2} O_{3} = \frac{475}{2} \times \frac{160}{1000} = 38 g$

$%$ $F e_{2} O_{3} = \frac{38}{38 + 9} \times 1 - 00 = 80.85 %$

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A mixture of FeO and Fe2O3 is completely reacted with 100 mL of 0.25M acidified KMnO4 solution. The resultant solution was then titrated with Zn dust which converted Fe3+ of the solution to Fe2+.The Fe2+ required 1000 mL of 0.10M K2Cr2O7 solution. Find out the weight %Fe2O3 in the mixture.