1.BeCl2+LiAlH4⟶X+LiCl+AlCl3\text{BeCl}_2 + \text{LiAlH}_4 \longrightarrow \text{X} + \text{LiCl} + \text{AlCl}_3BeCl2+LiAlH4⟶X+LiCl+AlCl3a.X is LiHb.X is BeH2\text{BeH}_2BeH2c.X is BeCl2⋅2H2O\text{BeCl}_2 \cdot 2\text{H}_2\text{O}BeCl2⋅2H2Od.NoneLogin to continueOnly logged in users canattempt or see the solution.
