When ferric oxide reacts with NaOH, the product formed is

NaF

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F e C l_{3}

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F e (O H)_{3}

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N a F e O_{2}

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Solution

The correct option is C $F e (O H)_{3}$
$F e_{2} O_{3} + 6 N a O H \to 2 F e (O H)_{3} B r o w n i n s o l u b l e i n N a O H ↓ + 3 N a_{2} O$

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