$BeCl_2$ reacts with $LiAlH_4$ to give ....
- A$Be + Li[AlCl_4] + H_2$
- B$Be + AlH_3 + LiCl + HCl$
- C$BeH_2 + LiCl + AlCl_3$
- D$BeH_2 + Li[AlCl_4]$
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