Question

1.0 mole of $Fe$ reacts completely with 0.65 mole of $O_2$ to give a mixture of only $FeO$ and $F{e}_2{O}_3$.
The mole ratio of ferrous oxide to ferric oxide is x: y, then (x+y) is

Answer 1

$\begin{array}{ccccc}2Fe& +& O_2& \to & 2FeO\\ \text{a mole}& & \frac{a}{2}\text{mole}& & a\\ 4Fe& +& 3O_2& \to & 2F{e}_2{O}_3\\ b& & \frac{3b}{4}& & \frac{b}{2}\end{array}$
$\therefore a+b=1$...(i)
$\therefore \frac{a}{2}+\frac{3b}{4}=0.65$...(ii)
On multiplying equation (ii) by 2, we get
$[\frac{a}{2}+\frac{3b}{4}=0.65]\times 2$
$a+\frac{3b}{2}=1.30$...(iii)
Subtracting equation (iii) from equation (i), we get
$[a+b=1]-[a+\frac{3b}{2}=1.30]$
$b=0.6\text{mole}$
Now, placing the value of $b=0.6$ in equation (i), we get,
$a=0.4$
$\therefore$ moles of $FeO=a=0.4$
moles of $F{e}_2{O}_3=\frac{b}{2}=0.3$
$\therefore$ ratio = 4 : 3
Now, $x+y=4+3=7$