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# When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C).

11.28    When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of $NaOH$ to give soluble complex (B). Compound (A) is 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.

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In given information, when X is treated with the sodium hydroxide it gives A(white ppt) and it is soluble in excess of $NaOH$. Thus X must be aluminium. So, the white ppt of A is aluminium hydroxide.
The complex B is soluble in excess of sodium hydroxide So it means it is a sodium tetrahydroxy aluminate(II) complex.

$\\2Al+3NaOH\rightarrow Al(OH)_3+3Na^+\\ Al(OH)_{3}+NaOH\rightarrow Na[Al(OH)_{3}]^-$

Now, when we add dilute HCl to aluminium hydroxide it gives the compound C (Aluminum chloride).

$Al(OH)_{3}+3HCl\rightarrow AlCl_{3}+3H_{2}O$
(A)                                           (C)

And when compound (A) is heated strongly, it gives D and this compound used to extract metal X

$2Al(OH)_{3}\rightarrow Al_2O_{3}+3H_{2}O$
(A)                          (D)

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