# A mixture of FeO and $Fe_3O_4$ when heated in air to a constant weight,gains 5% of its weight,Find the composition of the initial mixture.

$(a)\;21.5\%,78.5\%\qquad(b)\;22\%,78\%\qquad(c)\;20.25\%,79.75\%\qquad(d)\;20\%,80\%$

Let the initial wt of the mixture is 100gm then the final wt of the mixture after heating in air will be 105gm.
Let $x$ be the wt of $FeO$ in the initial mixture then wt of $Fe_3O_4=100-x$
When the mixture is heated in air.$(O_2)$
$4FeO+O_2\rightarrow 2Fe_2O_3$
$4Fe_3O_4+O_2\rightarrow 6Fe_2O_3$
4 moles $FeO$=2 moles of $Fe_2O_3$
$\Rightarrow \large\frac{x}{72}$ moles=$\large\frac{x}{144}$ moles of $Fe_2O_3$
4 moles $Fe_2O_4$=6 moles of $Fe_2O_3$
$\Rightarrow \large\frac{100-x}{232}$ moles=$\large\frac{6}{4}\big(\frac{100-x}{232}\big)$ moles of $Fe_2O_3$
Total moles of $Fe_2O_3=\large\frac{x}{144}+\frac{6}{4}\big(\large\frac{100-x}{232}\big)$
Wt of $Fe_2O_3=\big(\large\frac{x}{144}+\frac{6}{4}\big(\frac{100-x}{232})]$$\times 160=105$
$x=20.25gm$=wt of $FeO$
$\Rightarrow$ Wt of $Fe_3O_4=100-x$=79.75gm
% FeO=20.25