LiAlH4 / H2O reaction energy
So you have the reaction :

$$\ce{LiAlH4 + 4H2O &lt;=&gt; LiOH + Al(OH)3 + 4H2}$$

You want to get the "energy" given by this reaction, which is nothing but the reaction enthalpy. 

$\Delta_rH^0(T) = \sum_{i}\nu_i.\Delta H_{i,f}^0(T)$ 

Now you get this, just look for the theorical values of each $\Delta H_f^0$ you need for the 5 compounds you have (they are tabulated) and sum them up pondered with the stoechiometric coefficients.