Question

Of two oxides of iron, the first contained $22\%$ and the second contained $30\%$ of oxygen by weight. The ratio of weights of iron in the two oxides that combine with the same weight of oxygen, is

• A. $3:2$
• B. $2:1$
• C. $1:2$
• D. $1:1$

Answer 1

Answer: (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 $Fe=\frac{70}{56}=1.25$
Simpler molar ratio $=\frac{1.875}{1.25}=1.5,\frac{1.25}{1.25}=1$
$\therefore$ The formula of second oxide is $F{e}_2{O}_3$.
Suppose in both the oxides, iron reacts with $xg$ oxygen.
$\therefore$ Equivalent weight of $Fe$ in
$FeO$ or $=\frac{\text{ weight of }F{e}_{\text{II }}}{\text{ weight of oxygen }}\times 8$
$\frac{56}{2}=\frac{\text{ weight of }F{e}_{\text{II }}}{x}\times 8$
$\therefore$ Equivalent weight of $Fe$ in
$F{e}_2{O}_2=\frac{\text{ weight of }F{e}_{III}}{\text{ weight of oxygen }}\times 8$
$\frac{56}{3}=\frac{{\text{ weight of Fe }}_{\text{III }}}{x}\times 8.\text{..(ii) }$
From Eq. (i) and (ii),
$\frac{\text{ weight of}F{e}_{II}}{\text{ weight of}F{e}_{II}}=\frac{3}{2}$