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I'm dealing with a free-energy scam on another forum. It is well-known that $\ce{LiAlH4}$ reacts explosively with water. My chemistry days are long behind me, so I thought I'd come here to ask:

$1~\text{mol}$ of $\ce{LiAlH4}$ reacted with $4~\text{mol}$ of water gives how much energy?

I'm not about to do this, mind you. I was just curious about the specific numbers.

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So you have the reaction :

$$\ce{LiAlH4 + 4H2O <=> LiOH + Al(OH)3 + 4H2}$$

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.

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