A mixture of $F e O$ and $F e_{3} O_{4}$ when heated in air to a constant weight gains 55 in its mass. find the composition of the initial mixture.

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Solution

Let the % of $F e O$ in the mixture be $= x$
So, % of $F e_{3} O_{4}$ in the mixture $= (100 - x)$
$F e O$ on heating is converted into $F e_{2} O_{3}$ .

$4 F e O 288 g + O_{2} \to 2 F e_{2} O_{3} 320 g$

288 g of $F e O$ yield $= 320 g$ of $F e_{2} O_{3}$
$x g$ of $F e O$ will yield

$= \frac{320}{28} x g$ of $F e_{2} O_{3}$

$2 F e_{3} O_{4} 464 g + \frac{1}{2} O_{2} \to 3 F e_{2} O_{3} 480 g$

464 g of $F e_{3} O_{4}$ yield $= 480 g$ of $F e_{2} O_{3}$

(100-x) g of $F e_{2} O_{3}$ will yield $= \frac{480}{464} (100 - x)$ of $F e_{2} O_{3}$

Total $F e_{2} O_{3} = \frac{320}{288} x + \frac{480}{464} (100 - x)$

According to the question:

$\frac{320}{288} x + \frac{480}{464} (100 - x) = 105$

$x = 20.2$
So, percentage of $F e O = 20.2$
and percentage of $F e_{3} O_{4} = 79.8$

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