When excess of $N a O H$ is added to $A l C l_{3}$ solution, the product formed is:

A l_{2} O_{3}

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N a A l O_{2}

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N a_{3} A l O_{2}

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N a_{3} A l O_{3}

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Solution

The correct option is B $N a A l O_{2}$
When excess sodium hydroxide is added to aluminium chloride, sodium aluminate $(N a A l O_{2})$ is obtained.
$A l C l_{3} + 3 N a O H \to N a A l O_{2} + 2 H_{2} O$

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