BeCl2 reacts with LiAlH4 to give :
Solution
2BeCl<sub>2</sub> + LiAlH<sub>4</sub> $\to$ 2 BeH<sub>2</sub> + LiCl + AlCl<sub>3</sub><br/><br/>
The above reaction using LiAlH<sub>4</sub> is an important preparation method for production of hydrides.
About this question
Subject: Chemistry · Chapter: s-Block Elements · Topic: Alkali Metals
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