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NCERT Class XI Chemistry The p-Block Elements Solutions

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Question : 28 of 38
Marks: +1, -0
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.
Solution:  
The given data suggests that the metal X is aluminium.
Al(X)\underset{(X)}{\mathrm{Al}} + 2NaOH + 2H2O2\mathrm{H}_2\mathrm{O} → 2NaAlO2+3H22\mathrm{NaAlO}_2 + 3\mathrm{H}_2
NaAlO2+2H2O\mathrm{NaAlO}_2 + 2\mathrm{H}_2\mathrm{O} → Al(OH)3White gelationous ppt. (A)\underset{\text{White gelationous ppt. (A)}}{\mathrm{Al(OH)_3}} →NaOHExcess\xrightarrow[\mathrm{NaOH}]{\text{Excess}} Na[Al(OH)4]Sodium tetrahydroxo aluminate (III) (soluble , B)\underset{\text{Sodium tetrahydroxo aluminate (III) (soluble , B)}}{\mathrm{Na[Al(OH)_4]}}
Al(OH)3(A)+3HCl\underset{(A)}{\mathrm{Al(OH)_3}} + 3\mathrm{HCl} → AlCl3(C)+3H2O\underset{(C)}{\mathrm{AlCl_3}} + 3\mathrm{H}_2\mathrm{O}
2Al(OH)3(A)\underset{(A)}{2\mathrm{Al(OH)_3}} →Δ\xrightarrow{\Delta} Al2O3(D)+3H2O\underset{(D)}{\mathrm{Al_2O_3}} + 3\mathrm{H}_2\mathrm{O}
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