A mixture of $F e O$ and $F e_{3} O_{4}$ when heated in air to constant weight gains $5 %$ in its weight. The $%$ of $F e O$ in the sample is:(Give your answer upto one decimal only)

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Solution

When $F e O$ and $F e_{3} O_{4}$ are heated both change to $F e_{2} O_{3}$ .
Let weight of $F e O$ and $F e_{3} O_{4}$ be $x g$ and $y g$ .
Total weight of reactant = $(x + y) g$
Since weight increases $5 %$ on heating when $F e O$ and $F e_{3} O_{4}$ changes completely to $F e_{2} O_{3}$ .
Weight of $F e_{2} O_{3} = \frac{105}{100} (x + y) = 1.05 (x + y)$
$F e O x g m + F e_{3} O_{4} y g m ⟶ 2 F e_{2} O_{3} 1.05 (x + y)$
Applying POAC on $F e$ ,
$\frac{x}{72} + \frac{3 y}{232} = 2 \times \frac{1.05 (x + y)}{160}$
$∴ \frac{x}{y} = \frac{81}{319}$
% of $F e O = \frac{81}{81 + 319} \times = 20.25$ %
% of $F e_{3} O_{4} = 79.75$ %

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