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Mole Concept and Avogadro's Number

2.32 gm mixtu...

Question

2.32 gm mixture of FeO and $F e_{2} O_{3}$ when burnt in atmosphere of oxygen, 56 ml of $O_{2}$ is required at STP for complete oxidation. Find mole percentage of $F e_{2} O_{3}$ in the given mixture.

A

50%

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B

40%

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C

45%

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D

60%

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Solution

The correct option is D 50%
$2 F e O + \frac{1}{2} O_{2} \to F e_{2} O_{3}$
volume of $O_{2}$ at $S T P = 56 m l$
No. of mole of $O_{2} = \frac{56 \times 10^{- 3}}{224}$
$\frac{1}{2} m o l O_{2}$ oxidise $2 m o l F e O$
$2.5 \times 10^{- 3} m o l O - 2$ oxidise $\frac{2}{\frac{1}{2}} \times 25 \times 10^{- 3} m o l F e O$
man of $F e O = 1 \times 10^{- 2} \times (56 + 16)$
$= 0.72 g$
Mass of $F e O_{3} = 2.32 - 0.72$
$= 1.6 g$
number of mole of $F e O_{3} = \frac{1.6}{(56 \times 2 + 3 \times 16)}$
$= 0.01$
mole $%$ of $F e O_{3} = \frac{0.01}{0.01 + 0.01} \times 100$
$= 50 %$

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