Question

If a mixture of $FeO$ and $F{e}_3{O}_4$ contains $75\%Fe,$ what will be the percentage amount of each oxide in the mixture?

• A. $64.10\%FeO$ and $35.90\%F{e}_2{O}_3$
• B. $50\%FeO$ and $50\%F{e}_2{O}_3$
• C. $75\%FeO$ and $25\%F{e}_3{O}_4$
• D. $35.90\%FeO$ and $64.10\%F{e}_2{O}_3$

Answer 1

Answer: (A)

A mixture of $FeO$ and $F{e}_3{O}_4$ (i.e. $FeO\cdot F{e}_2{O}_3$ )

actually consists of 2 units of $FeO$ and 1 unit of $F{e}_2{O}_3.$ 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 $F{e}_2{O}_3$

$=\frac{(\text{ Atomic weight of Fe })\times 2}{\text{ Formula weight of }F{e}_2{O}_3}\times 100$

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

Suppose, ' $x^{\prime }g$ of $FeO$ (in total) and ' $y^{\prime }g$ of $F{e}_2{O}_3$

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 $F{c}_2{O}_3$ in mixture ( $y$ )

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