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Metal M is treated with NaOH, a white gelatinous precipitate X is formed. X is soluble in excess of NaOH solution. X when heated strongly gives an oxide, which is also one of the products of alumino thermite process. The metal M is
X is $$\text{Al(OH)}_3$$, a white gelatinous precipitate that dissolves in excess NaOH as sodium aluminate. On strong heating it gives $$\text{Al}_2\text{O}_3$$, which is formed in the thermite reaction $$\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$$. So M is Al.
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