A silvery-white metal on treatment with $N a O H$ and $H C l$ liberates $H_{2}$ gas to form B and C respectively. The metal A will not react with acid D due to the formation of a passive film on the surface. Hence, it is used for transporting acid D. Identify A, B, C, D and support your answer with balanced equations.

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Solution

The silvery-white metal that reacts with both acids and bases is $A l$ .
$2 A l + 6 H C l ⟶ 2 A l C l_{3} + 3 H_{2} ↑$
$2 A l + 2 N a O H + 6 H_{2} O ⟶ 2 N a^{+} + 2 [A l (O H)_{4}]^{-} + 3 H_{2} ↑$
However $A l$ does not react with concentrated $H N O_{3}$ due to formation of passive oxide layer on the surface.
Hence, A= $A l$ , B= $N a [A l (O H)_{4}]$ , C= $A l C l_{3}$ , D= $H N O_{3}$ (conc.)

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An organic compound A (molecular formula $C_{2} H_{4} O_{2})$ reacts with Na metal to form a compound B and evolves a gas which burns with a pop sound. Compound A on treatment with an alcohol C in the presence of a little of concentrated sulphuric acid forms a sweet-smelling compound D (molecular formula $C_{2} H_{6} O_{2})$ . Compound D on treatment with NaOH solution gives back B and C. Indentity A, B, C and D.