Ethyl chloride $\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}\right)$is prepared by reaction of ethylene With hydrogen chloride:

$$\begin{gathered} {\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+\mathrm{HCl}\left(\mathrm{g}\right)⟶{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}\left(\mathrm{g}\right) \\ \mathrm{\Delta }H=-72.3\mathrm{kJ}/\mathrm{mol}\text{. } \end{gathered}$$

What is the value of $\mathrm{\Delta }E$ (in KJ) if 98 g of ethylene and 109.5 g of HCl are allowed to react at 300 K ?