How many moles of ${\mathrm{Fe}}^{2+}$ ions are formed when excess of iron react with $500 \mathrm{mL} \mathrm{O}.4 \mathrm{N} \mathrm{HCl}$, under inert atmosphere?

(assume no change in volume)

$\because$ Reaction between $\mathrm{Fe}$ and $\mathrm{HCl}$ is as follows

$\mathrm{Fe}+2\mathrm{HCl}⟶\underset{(+2)}{{\mathrm{FeCl}}_2}+{\mathrm{H}}_2$ on reaction

$\mathrm{n}$(moles of $\mathrm{Fe}$) $=\mathrm{n}$ (moles of $\mathrm{HCl}$)

$\mathrm{n}\left(\mathrm{Fe}\right)=$ Normality $\left(\mathrm{HCl}\right)\times$ Volume of $\mathrm{HCl}$

$\mathrm{n}\left(\mathrm{Fe}\right)=\frac{0.4\times 500}{1000}=0.2 \mathrm{moles}$

As each $\mathrm{Fe}$ consumers $2$ electrons to change to ${\mathrm{Fe}}^{2+}$ ion.

$\therefore$ Number of ${\mathrm{Fe}}^{2+}$ ions produced $=\frac{0.2}{2}=0.1 \mathrm{mole}$