Chemistry - Mole Concept Question with Solution | TestHub

$\mathrm{F}{\mathrm{e}}_{0.95}{\mathrm{O}}_1$
Let there are x ions of$\mathrm{F}{\mathrm{e}}^{3+}$then$\left(0.95-\mathrm{x}\right)$ions will be$\mathrm{F}{\mathrm{e}}^{2+}$. Compound is electrically neutral.
$+3\mathrm{x}+2(0.95-\mathrm{x})-2=0$
$3\mathrm{x}+1.90-2\mathrm{x}-2=0$
$\mathrm{x}=0.1$
$\mathrm{\%}\mathrm{F}{\mathrm{e}}^{3+}=\frac{0.1}{0.95}\times 100$
$=10.5$