The number of moles of Fe taken

$=\frac{112}{56}=2$

So the chemical reaction is

$2\mathrm{F}\mathrm{e}\left(\mathrm{s}\right)+4\mathrm{H}\mathrm{C}\mathrm{l}\left(\mathrm{a}\mathrm{q}\right)\to 2\mathrm{F}\mathrm{e}\mathrm{C}{\mathrm{l}}_2\left(\mathrm{a}\mathrm{q}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)$

New work done $=\mathrm{P}\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)$

Since ${\mathrm{V}}_2-{\mathrm{V}}_1={\mathrm{V}}_{{\mathrm{H}}_2}$

Hence volume of solids and liquids is taken as $\cong 0$

So work done $=\mathrm{P}.{\mathrm{V}}_{{\mathrm{H}}_2}$

also $\mathrm{P}{\mathrm{V}}_{{\mathrm{H}}_2}=\mathrm{n}\mathrm{R}\mathrm{T}$ or ${\mathrm{V}}_{{\mathrm{H}}_2}=\frac{\mathrm{n}\mathrm{R}\mathrm{T}}{\mathrm{P}}$

So work done $=\mathrm{P}.\frac{\mathrm{n}\mathrm{R}\mathrm{T}}{\mathrm{P}}=\mathrm{n}\mathrm{R}\mathrm{T}$ [ $\mathrm{R}=2$ Cal]

$=2\times 2\times 300=1200$ Cal.