`BeCl_(2) + LiAlH_(4)`

1.	`BeCl_(2) + LiAlH_(4)`
Answer» `underset({:("Beryllium"),("chloride"):})(2BeCl_(2)) + underset({:("Lithium"),("aluminium"),("hydride"):})(LiAIH_(4)) to 2BeH_(2) + LiCI + AICI_(3)`

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