(BeCl\_2) reacts with (LiAlH\_4) to give: | Chemistry | SolveeIt

$BeCl_2$ reacts with $LiAlH_4$ to give:

$Be + Li[AlCI_4] + H_2$

$Be + AlH_3 + LiCI + HCI$

$BeH_2 + LiCl + AlCl_3$

$BeH_2 + Li[AlCl_4]$

Answer

$BeH_2 + LiCl + AlCl_3$

Explanation

$BeCl_2$ reacts with $LiAlH_4$ to give the following chemical reaction

$BeCl_2+LiAlH_4 → BeH_2+LiCl+Alcl_3$

Chemistry Question on Chemical Reactions