🛠️ JEE🧪 Chemistry

\(\mathrm{Al}_2 \mathrm{O}_3\) was leached with alkali to get \(\mathrm{X}\). The solution of X on passing of gas \(\mat…

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\(\mathrm{Al}_2 \mathrm{O}_3\) was leached with alkali to get \(\mathrm{X}\). The solution of X on passing of gas \(\mathrm{Y}\), forms \(\mathrm{Z}\). X, Y and \(Z\), respectively, are

a

\(\mathrm{X}=\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right], \mathrm{Y}=\mathrm{SO}_2, \mathrm{Z}=\mathrm{Al}_2 \mathrm{O}_3\)

b

\(\mathrm{X}=\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right],\mathrm{Y}=\mathrm{CO}_2, \mathrm{Z}=\) \(\mathrm{Al}_2 \mathrm{O}_3 \cdot \mathrm{xH}_2 \mathrm{O}\)

c

\(\mathrm{X}=\mathrm{Al}(\mathrm{OH})_3, \mathrm{Y}=\mathrm{CO}_2, \mathrm{Z}=\mathrm{Al}_2 \mathrm{O}_3\)

d

\(\mathrm{X}=\mathrm{Al}(\mathrm{OH})_3, \mathrm{Y}=\mathrm{SO}_2, \mathrm{Z}=\mathrm{Al}_2 \mathrm{O}_3 \cdot \mathrm{xH}_2 \mathrm{O}\)

✓ Correct answer: b)

\(\mathrm{X}=\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right],\mathrm{Y}=\mathrm{CO}_2, \mathrm{Z}=\) \(\mathrm{Al}_2 \mathrm{O}_3 \cdot \mathrm{xH}_2 \mathrm{O}\)

Explanation

When bauxite (\(Al_{2}O_{3}\)) is leached with a hot, concentrated solution of sodium hydroxide (\(NaOH\)), it dissolves to form soluble sodium tetrahydroxoaluminate(III) (or sodium aluminate)
The aluminate solution is then neutralized by passing carbon dioxide (\(CO_{2}\)) gas through it. This induces the precipitation of hydrated aluminum oxide.
\({\mathrm{Al}}_{2}{\mathrm{O}}_{3}(\mathrm{s})+2\mathrm{NaOH}(\mathrm{aq})+3{\mathrm{H}}_{2}\mathrm{O}(\mathrm{l})\to 2\mathrm{Na}[\mathrm{Al}{(\mathrm{OH})}_{4}](\mathrm{aq})\\ 2\mathrm{Na}[\mathrm{Al}{(\mathrm{OH})}_{4}](\mathrm{aq})+{\mathrm{CO}}_{2}(\mathrm{g})\to {\mathrm{Al}}_{2}{\mathrm{O}}_{3}.{\mathrm{xH}}_{2}\mathrm{O}(\mathrm{s})+2{\mathrm{NaHCO}}_{3}(\mathrm{aq})\\ \mathrm{X}=\mathrm{Na}[\mathrm{Al}{(\mathrm{OH})}_{4}](\mathrm{aq}),\mathrm{Y}={\mathrm{CO}}_{2}(\mathrm{g}),\mathrm{Z}={\mathrm{Al}}_{2}{\mathrm{O}}_{3}.{\mathrm{xH}}_{2}\mathrm{O}(\mathrm{s})\)

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