In the leaching of alumina from bauxite, the ore expected to leach out in the process by reacting with $N a O H$ is

T i O_{2}

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Z n O

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S i O_{2}

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F e_{2} O_{3}

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Solution

The correct option is C $S i O_{2}$
Along with $A l_{2} O_{3}$ , $S i O_{2}$ too dissolves forming $N a S i O_{3}$ .

In bauxite impurities of $F e_{2} O_{3}$ , $T i O_{2}$ and $S i O_{2}$ are present, $F e 2 O_{3}$ and $T i O_{2}$ are basic oxides therefore does not reacts with or dissolve in $N a O H$ whereas $S i O_{2}$ is acidic oxide it gets dissolve in $N a O H$ , hence leach out

$S i O_{2} + 2 N a O H \to N a_{2} S i O_{3} (a q .) + H_{2} O$

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Similar questions

In Bayer’s leaching process, the siliceous impurities $(S i O_{2})$ are filtered out immediately after leaching with NaOH – True or False?

A l_{2} O_{3}

N a O H

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