A mixture of $F e O$ and ${F e}_{3} O_{4}$ heated in air to a constant weight gains $5$ % in its weight. The percentage contribution of $F e O$ in the mixture is:

20.25

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79.75

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28.13

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71.87

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Solution

The correct option is A $20.25$ %
Total weight of mixture $= 100 g$

$F e O = x, {F e}_{3} O_{4} = 100 - x$

constant weight on heating $= 105 g$

It implies chemically, the conversion of:

$2 F e O 2 m o l e s ⟶ {F e}_{2} O_{3} 1 m o l e$

and ${F e}_{3} O_{4} ⟶ \frac{3}{2} {F e}_{2} O_{3}$
$x g F e O = \frac{x}{72} m o l \Rightarrow \frac{x}{72} \times \frac{1}{2} m o l {F e}_{2} O_{3}$

$= \frac{x \times 100}{72 \times 2} g o f {F e}_{2} O_{3} = \frac{10 x}{9}$

$(100 - x) g {F e}_{3} O_{4} = \frac{(100 - x)}{232} m o l \Rightarrow \frac{(100 - x)}{232} \times \frac{3}{2} m o l {F e}_{2} O_{3} = \frac{30}{29} (100 - x) g {F e}_{2} O_{3}$

Thus $\frac{10 x}{9} + \frac{30}{9} (100 - x) = 105$

On solving, $x = 20.25$ %

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