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A 3.0 g sam...

Question

A $3.0 g$ sample containing $F e_{3} O_{4}, F e_{2} O_{3}$ and an inert impure substance is treated with excess of $K I$ solution in presence of dilute $H_{2} S O_{4}$ . The entire iron is converted to $F e^{2 +}$ along with the liberation of iodine. The resulting solution is diluted to $100 m L$ . A $20 m L$ of the dilute solution requires $11.0 m L$ of $0.5 M N a_{2} S_{2} O_{3}$ solution to reduce the iodine present. A $50 m L$ of the diluted solution after complete extraction of iodine requires $12.8 m L$ of $0.25 M K M n O_{4}$ solution in dilute $H_{2} S O_{4}$ medium for oxidation of $F e^{2 +}$ . Calculate the percentage of $F e_{2} O_{3}$ and $F e_{3} O_{4}$ in the original sample.

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Solution

$F e_{3} O_{4}$ is an equimolar mixture of $F e_{2} O_{3}$ and $F e O$ . Thus, the sample contains $F e_{2} O_{3}, F e O$ and impurities. The amount of iodine libeated depends on the amount of $F e_{2} O_{3}$ and the entire iron is converted into $F e^{2 +}$ .
$F e_{3} O_{4} + 2 K I + H_{2} S O_{4} \to 3 F e O + H_{2} O + K_{2} S O_{4} + I_{2}$
$F e_{2} O_{3} + K I + H_{2} S O_{4} \to 2 F e O + H_{2} O + K_{2} S O_{4} + I_{2}$
$5 \times 11.0 m L$ of $0.5 M N a_{2} S_{2} O_{3} \equiv 55.0 m L$ of $0.5 N N a_{2} S_{2} O_{3} s o l n$ .
$\equiv 55.0 m L$ of $0.5 N I_{2} s o l n .$
$\equiv 55.0 m L$ of $0.5 N F e_{2} O_{3} s o l n .$
$= 27.5 \times 10^{- 3}$ equivalent $F e_{3} O_{4}$ soln.
$= 13.75 \times 10^{- 3} m o l e s F e_{2} O_{3}$
$2 \times 12.8 m L$ of $0.25 M K M n O_{4} s o l n .$
$\equiv 25.6 m L$ of $1.25 N K M n O_{4} s o l n .$
$\equiv 25.6 m L$ of $1.25 N F e O s o l n .$
$= 32.0 \times 10^{- 3}$ equivalent $F e O$
$= 32.0 \times 10^{- 3} m o l e s F e O$
Moles of $F e O$ in $F e_{2} O_{4} = 0.032 - 0.0275 = 0.0045$
Mass of $F e_{3} O_{4} = 0.0045 \times 232 = 1.044 g$
Moles of $F e_{2} O_{3}$ existing separately
$= 0.01375 - 0.0045 = 0.00925$
% $F e_{3} O_{4} = \frac{1.044}{3} \times 100 = 34.8$
% $F e_{2} O_{3} = \frac{1.48}{3} \times 100 = 49.33$ .

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