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 $F e=\frac{78}{56}=1.392$
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 $F e=\frac{70}{56}=1.25$
Simpler molar ratio $=\frac{1.875}{1.25}=1.5, \frac{1.25}{1.25}=1$
$\therefore $ The formula of second oxide is $Fe _{2} O _{3}$.
Suppose in both the oxides, iron reacts with $x g$ oxygen.
$\therefore $ Equivalent weight of $Fe$ in
$FeO$ or $=\frac{\text { weight of } Fe _{\text {II }}}{\text { weight of oxygen }} \times 8$
$\frac{56}{2}=\frac{\text { weight of } Fe _{\text {II }}}{x} \times 8$
$\therefore $ Equivalent weight of $Fe$ in
$Fe _{2} O _{2}=\frac{\text { weight of } Fe _{ III }}{\text { weight of oxygen }} \times 8$
$\frac{56}{3}=\frac{\text { weight of Fe }_{\text {III }}}{x} \times 8 . \text {..(ii) }$
From Eq. (i) and (ii),
$\frac{\text { weight of} Fe_{II}}{\text { weight of} Fe_{II}}=\frac{3}{2}$