Write a balanced chemical equation when sodium iodate is added to a solution of sodium bisulphite.

2 N a I O_{3} + 5 N a H S O_{3} \to 3 N a H S O_{4} + 2 N a_{2} S O_{4} + I_{2} + H_{2} O

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2 N a I O_{3} + 5 N a H S O_{3} \to 5 N a H S O_{4} + H_{2} S O_{4} + I_{2} + H_{2} O

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2 N a I O_{3} + 5 N a H S O_{3} \to 3 N a H S O_{4} + 2 N a_{2} S O_{4} + H I O_{3} + H_{2} O

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None of these

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Solution

The correct option is B $2 N a I O_{3} + 5 N a H S O_{3} \to 3 N a H S O_{4} + 2 N a_{2} S O_{4} + I_{2} + H_{2} O$
Sodium iodate is a strong oxidizing agent. When sodium iodate is added to a solution of sodium bisulphite, it releases iodine and oxidizes ${H S O_{3}}^{-}$ to ${H S O_{4}}^{-}$ and ${S O_{4}}^{2 -}$ .

The balance chemical equation is:

$2 N a I O_{3} + 5 N a H S O_{3} \to 3 N a H S O_{4} + 2 N a_{2} S O_{4} + I_{2} + H_{2} O$

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Similar questions

2 N a I O_{3} + 5 N a H S O_{3} \to 2 N a H S O_{4} + 2 N a_{2} S O_{4} + H_{2} O + I_{2}

N a H S O_{3}

381

I_{2}

C a C O_{3} + H C l \to C a C l_{2} + H_{2} O + C O_{2}

M n O_{2} + H C l \to M n C l_{2} + H_{2} O + C l_{2}

M g (O H)_{2} + H C l \to M g C l_{2} + H_{2} O

K I + H_{2} S O_{4} \to K H S O_{4} + I_{2} + S O_{2}

I^{-} + H_{2} O_{2} \to H_{2} O + I_{2}

K I + H_{2} S O_{4} \to K_{2} S O_{4} + I_{2} + S O_{2} + H_{2} O

H_{2} {S O}_{4}

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