Question:
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.
Answer:
When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, So the metal X should be Al. The precipitate, which is obtained on treating with NaOH is Al (OH)3. So compound (A) is Al (OH)3.
When Al (OH)3 is treated with more of NaOH, it dissolves to form compound (B) .The compound (B) should be NaAlO2.
Al (OH)3 being amphoteric in nature reacts with HCl to form AlCl3 . So Compound (C) is AlCl3.
(A) on heating gives (D) which is used to extract metals.(D) should be Al2O3. The reactions are as given as -
2Al +3NaOH → Al (OH)3 + Na+
Al (OH)3 + NaOH → NaAlO2 + 2H2O
Al (OH)3 + 3HCl → AlCl3 + H2O
2Al (OH)3 (heat) → Al2O3 + 3H2O
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