When ${C l}_{2}$ is passed through cold dil. $N a O H$ , the products are :

N a C l, N a O C l a n d H_{2} O

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N a C l, {N a C l}_{2} a n d H_{2} O

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N a C l, {N a C l}_{3} a n d H_{2} O

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N a C l, {N a C l}_{4} a n d H_{2} O

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Solution

The correct option is A $N a C l, N a O C l a n d H_{2} O$
When ${C l}_{2}$ is passed through cold dil. $N a O H$ , the reaction is:

$C l_{2} + 2 N a O H \to N a C l + N a O C l + H_{2} O$

So, option A is correct.

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Balance

NaCl + H2O =NaOH + H2O + Cl2

N a C l

H_{2} O

N a O H +___\to N a C l O_{2} + H_{2} O

N H_{3}

N H_{4} C l + N a C l

N a O H

N H_{4} C l + N a O H \to N H_{3} + N a C l + H_{2} O

${C O}_{2} + N a O H \to H_{2} O + X$

Identify X.

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