Question

If a mixture of $FeO$ and $Fe_3O_4$ contains $75\% Fe,$ what will be the percentage amount of each oxide in the mixture?

• A. $64.10\% FeO$ and $35.90\% Fe_2O_3$
• B. $50\% FeO$ and $50\% Fe_2O_3$
• C. $75\% FeO$ and $25\% Fe_3O_4$
• D. $35.90\% FeO$ and $64.10\% Fe_2O_3$

Answer 1

Answer: (A)

A mixture of $FeO$ and $Fe _{3} O _{4}$ (i.e. $FeO \cdot Fe _{2} O _{3}$ )

actually consists of 2 units of $FeO$ and 1 unit of $Fe _{2} O _{3} .$ In case of $FeO$,

The percentage of $Fe$ in $FeO$

$=\frac{\text { Atomic weight of } Fe }{\text { Formula weight of } FeO } \times 100$

$=\frac{56}{72} \times 100=77.78 \%$

The percentage of $Fe$ in $Fe _{2} O _{3}$

$=\frac{(\text { Atomic weight of Fe }) \times 2}{\text { Formula weight of } Fe _{2} O _{3}} \times 100$

$=\frac{112}{160} \times 100=70.00 \%$

Suppose, ' $x'g$ of $FeO$ (in total) and ' $y' g$ of $Fe _{2} O _{3}$

form the mixture that contains $75 \%$ Fe.

Hence mathematically,

$77.78 \%$ of $x+70.00 \% y=75 \%$ of $(x +y)$

or $77.78 \%$ of $x+70.00 \% y=75.00$ of $x$

$+75.00 \% $ of $y$

or $(77.78-75.00 \%)$ of $x=(75.00-70.00) \%$ of $y$

or $2.77 \%$ of $x=5.00 \%$ of $y$

or $\frac{ x }{ y }=\frac{5.00 \%}{2.78}=\frac{500}{278}=\frac{250}{139}$

or $x: y=250: 139$

Now, percent amount of FeO in mixture (x)

$=\frac{250}{(250+139)} \times 100=\frac{25000}{389} \%=64.3 \%$

Hence, percent amount of $Fc _{2} O _{3}$ in mixture ( $y$ )

$=(100-64.3) \%=35.7 \%$