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When metal X is treated with sodium hydr...

When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is solubel in excess of NaOH to give soluble complex (B). Compound (A) si soluble in dilute HCl to form compound (C ). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C ) and (D). Write suitable equations to support their identities.

Text Solution

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As metal X, on treatment with NaOH forms a white precipitate, which dissolves in excess NaOH to form soluble complex B, the metal X is Al.
`underset((X))(Al) + 3NaOH to underset("white (A)")(Al(OH)_3)darr + 3Na^(+)`
`{:(underset((A))(Al(OH)_3)+ NaOH("excess")underset((B))(to2Na^(+))[Al(OH)_4]^(-)),(" Sodium tetrahydroxoaluminate(III)"):}`
`underset((A))(Al(OH)_3) + 3HCl (aq) to underset(C )(AlCl_3)+ 3H_2O`
`underset((A))(2Al(OH)_3)to underset((D))(AlCl_3) + 3H_2O`
Thus, X = Al (aluminium), `A = (Al(OH)_3)`(aluminium hydroxide), `B = Na[Al(OH)_4]" " C = AlCl_3` (aluminium chloride), `D = Al_2O_3` (alumina)
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