Question

A $3:2$ molar ratio mixture of $FeO$ and $F{e}_2{O}_3$ react with oxygen to produce a $2:3$ molar ratio mixture of $FeO$ and $F{e}_2{O}_3$. Find the mass (in g) of $O_2$ gas required per mole of the initial mixture.

Answer 1

From one - mole of initial mixture, some $FeO$ must have reacted with oxygen and got converted into $F{e}_2{O}_3$.
$4FeO+O_2\to 2F{e}_2{O}_3$
Initial moles $\frac{3}{5}\frac{2}{5}$
Final moles $\frac{3}{5}-x\frac{2}{5}+\frac{x}{2}$

But, ratio of final moles is $2:3$.

$\therefore \frac{(\frac{3}{5}-x)}{(\frac{2}{5}+\frac{x}{2})}=\frac{2}{3}$
$\Rightarrow 3(\frac{3}{5}-x)=2(\frac{2}{5}+\frac{x}{2})$
$\therefore x=\frac{1}{4}$
$\therefore$ Moles of $FeO$ reacted $=x=\frac{1}{4}$
$\therefore$ Moles of $O_2$ required $=\frac{1}{4}\left(x\right)=\frac{1}{16}$
$\therefore$ Mass of $O_2$ required $=\frac{1}{16}\times 32=2g$