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.

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

The correct option is A 2

$\begin{array}{c}\text{From one mole of initial mixture,}\\ \text{Some FeO must have reacted with}\\ \text{oxygen and got converted into}{Fe}_2{O}_3\text{.}\end{array}$

$\begin{array}{c}4FeO+O_2\to 2{Fe}_2{O}_3\\ 4FeO+O_2\to 2{Fe}_2{O}_3\end{array}$

$\begin{array}{c}\text{Initial}\text{moles}\frac{3}{5}\left(FeO\right)\frac{2}{5}\left(F{e}_2{O}_3\right)\\ \text{Final mole}\frac{3}{5}-x\left(FeO\right)\frac{2}{5}+\frac{x}{2}\left(F{e}_2{O}_3\right)\end{array}$

$\begin{array}{c}\text{But final ratio is}2\text{: 3}\\ \begin{array}{c}\frac{(\frac{3}{5}-x)}{(\frac{2}{5}+\frac{x}{2})}=\frac{2}{3}\\ x=\frac{1}{4}\end{array}\end{array}$

$\begin{array}{c}\text{Moles of FeO reacted}=\frac{1}{4}\\ \text{Moles of}{O}_2\text{required}=\frac{1}{4}x=0.0625\\ \text{Mass of}{O}_2\text{required}=0.0625\times 32=2g\\ \end{array}$