Question

A mixture of ferrous oxide and ferric oxide contains $75\%$. The percentage amount of ferrous oxide in the mixture is -----
(Atomic mass of $Fe = 56$)
(A) $53.10$
(B) $64.10$
(C) $33.42$
(D) $78.12$

Answer 1

The chemical formula for ferrous oxide is $FeO$, and that of ferric oxide is $Fe_2O_3$. The percentage weight of $FeO$ in the mixture can be calculated by finding the weight of $FeO$ in the same.

Complete step by step answer:
Let the weight of the total mixture of ferrous oxide and ferric oxide be taken as $100\ g$.
Let the amount of $FeO$ present in the mixture be $x\ g$, and the amount of $Fe_2O_3$ present in the mixture be $(100-x)\ g$.
The percentage of $Fe$ in the entire mixture is given as $75\%$. So, we will proceed by calculating the weight of $Fe$ in the entire mixture.
$FeO$:
Number of moles of $FeO$ present in the mixture $= \dfrac{weight\ of\ FeO}{molecular\ weight\ of\ FeO}$
Now, the molecular weight of $FeO=$ Atomic weight of $Fe$ $+$ Atomic weight of $O$
$\Rightarrow$ Molecular weight of $FeO = 56\ g + 16\ g = 72\ g$
Therefore,
Number of moles of $FeO$ present in the mixture $= \dfrac{x\ g}{72\ g} = \dfrac{x}{72}$
If one mole of $FeO$ contains $56\ g$ of $Fe$, then $\dfrac{x}{72}$ moles of $FeO$ will contain $\dfrac{x}{72}\times 56\ g$ of $Fe$.
$Fe_2O_3$:
Number of moles of $Fe_2O_3$ present in the mixture $= \dfrac{Weight\ of\ Fe_2O_3}{Molecular\ weight\ of\ Fe_2O_3}$
Now, the molecular weight of $Fe_2O_3$ $=$ $2\times$ Atomic weight of $Fe$ $+$ $3\times$ Atomic weight of $O$
$\Rightarrow$ Molecular weight of $Fe_2O_3 = 2\times 56\ g + 3\times 16\ g = 112\ g + 48\ g = 160\ g$
Therefore,
The number of moles of $Fe_2O_3$ present in the mixture $= \dfrac{(100-x)\ g}{160\ g} = \dfrac{100-x}{160}$
If one mole of $Fe_2O_3$ contains $2\times 56\ g$ of $Fe$, then $\dfrac{100-x}{160}$ moles of $Fe_2O_3$ will contain $\dfrac{(100-x) }{160}\times 2\times 56\ g$ of $Fe$.
The total amount of $Fe$ present in the mixture $=$ Amount of $Fe$ present in $FeO$ $+$ Amount of $Fe$ present in $Fe_2O_3$
$\Rightarrow$ Total amount of $Fe$ present in the mixture $=\dfrac{x}{72}\times 56 + \dfrac{(100-x)\times 2\times 56}{160}$
According to the question,
$\dfrac{x}{72}\times 56 + \dfrac{(100-x)\times 2\times 56}{160} = 75\%$
$\Rightarrow \dfrac{56x}{72} + \dfrac{(100-x)\times 112}{160} = \dfrac{75}{100}\times 100$
$\Rightarrow 0.778x + 0.7(100 – x) = 75$
$\Rightarrow 0.778x + 70 – 0.7x = 75$
$\Rightarrow 0.078x = 75 - 70$
$\Rightarrow 0.078x = 5$
$\Rightarrow x = \dfrac{5}{0.078}$
$\therefore x = 64.10$

So, the correct answer is Option B.

Note: 1. The total weight of the mixture is assumed to be $100\ g$ for convenience in calculations.
2. While calculating the molecular weight of a compound, the atomic weight of each atom present in the compound should be multiplied by the number of that atom present in it.