Explanation

(a)
For first oxide,
Moles of oxygen=$\frac{22}{16} = 1.375$,
Moles of Fe$= \frac{78}{56} = 1.392$
Simpler molar ratio, $\frac{1.375}{1.375} = 1,\ \frac{1.392}{1.375} = 1$
$\therefore$ The formula of first oxide is FeO.
Similarly for second oxide,
Moles of oxygen=$\frac{30}{16} = 1.875$,
Moles of Fe=$\frac{70}{56} = 1.25$
Simple molar ratio=$\frac{1.875}{1.25} = 1.5,\ \frac{1.25}{1.25} = 1$
$\therefore$ The formula of second oxide is $Fe_{2}O_{3}$.
Suppose in both the oxides, iron reacts with $x$g of oxygen.
$\therefore$ Equivalent weight of Fe in FeO
$\frac{weight\ of\ Fe_{II}\ }{weight\ of\ oxygen} \times 8$
$\frac{56}{2} = \frac{weight\ of\ Fe_{II}}{x} \times 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$\therefore$ Equivalent weight of Fe in $Fe_{2}O_{3}$
$= \frac{weight\ of\ Fe_{III}}{weight\ of\ oxygen} \times 8$
$\frac{56}{3} = \frac{weight\ of\ Fe_{III}}{x} \times 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eq. (i) and (ii),
$\frac{weight\ of\ Fe_{II}}{weight\ of\ Fe_{III}} = \frac{3}{2}$