Question

A mixture of $FeO$ and $Fe _{3} O _{4}$ heated in air to a constant weight gains $5 \%$ in its weight calculate composition of the mixture.

Answer 1

SOLUTION : We assume that total weight of the mixture $=100 g$ and wt. of $FeO =x$ g. Hence wt. of $Fe _{3} O _{4}=(100-x) g$

On heating to a constant weight, the wt. $=105 g$

Now $x g FeO =\frac{x}{72}$ mole $FeO =\frac{x}{2 \times 72}$ mole $Fe _{2} O _{3}=\frac{x \times 160}{2 \times 72} g Fe _{2} O _{3}=\frac{10 x}{9}$ $(100-x) g Fe _{3} O _{4}=\frac{100-x}{232}$ mole $Fe _{3} O _{4}=\frac{100-x}{232} \times \frac{3}{2}$ mole $Fe _{2} O _{3}$

$=\frac{100-x}{232} \times \frac{3}{2} \times 160 g =\frac{30}{29}(100-x) g$

Now $\quad \frac{10 x}{9}+\frac{30}{29}(100-x)=105, \frac{x}{9}-\frac{3 x}{29}+\frac{300}{29}=10 \cdot 5$

or, $\quad \frac{2 x}{9 \times 29}=\frac{304 \cdot 5-300}{29}=\frac{4 \cdot 5}{29}$

$\therefore \quad x=\frac{4 \cdot 5 \times 9}{2}=20 \cdot 25 \% FeO ;$ and $79 \cdot 75 \% Fe _{3} O _{4}$.