Lithium aluminium hydride can be prepared from the reaction of:
\(LiCl \ and \ Al_2H_6\)
\(LiH \ and \ Al_2Cl_6\)
\(LiCl, Al \ and \ H_2\)
\(LiH \ and \ Al(OH)_3\)
Step 1: Reaction for Preparation of \(\text{LiAlH}_4\)
Lithium aluminium hydride \(\text{LiAlH}_4\) is prepared by the reaction of lithium hydride (\(\text{LiH}\)) with aluminium chloride \(\text{Al}_2\text{Cl}_6\). The reaction is as follows:
\[8\text{LiH} + \text{Al}_2\text{Cl}_6 \rightarrow 2\text{LiAlH}_4 + 6\text{LiCl}.\]
Conclusion: The correct reactants for preparing \(\text{LiAlH}_4\) are \(\text{LiH}\) and \(\text{Al}_2\text{Cl}_6\). Therefore, the correct answer is (2)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: