In the following reactions- product A and B- respectively- are-NaOH - Cl 2- A - side productshot - conc-Ca OH 2- Cl 2- B - side productsdryA- NaClO 3 and Ca ClO 32B- NaOCl and Ca OCl 2C- NaOCl and Ca ClO 32D- NaClO 3 and Ca OCl 2

The correct option is D NaClO_3 and Ca(OCl)_26NaOH + 3Cl_2 → 5NaCl + NaClO_3 + 3H_2O 2Ca(OH)_2 + Cl_2 → Ca(OCl)_2 + CaCl_2 + H_2 O ...

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Carbon Dioxide

In the follow...

Question

In the following reactions, product (A) and (B), respectively, are:
$N a O H + C l_{2} \to (A) + side products$
(hot & conc.)

$C a (O H)_{2} + C l_{2} \to (B) + side products$
(dry)

N a C l O_{3}

and

C a (C l O_{3})_{2}

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N a O C l

and

C a (C l O_{3})_{2}

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N a O C l

and

C a (O C l)_{2}

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N a C l O_{3}

and

C a (O C l)_{2}

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Similar questions

+ C l_{2} \to N a C l + N a C l O_{3}

C l_{2}

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

3 N a C l O \to 2 N a C l + N a C l O_{3}

4 N a C l O_{3} \to 3 N a C l O_{4} + N a C l

C l_{2}

106.5

N a C l O_{3}

2 K I + A 2 \to K C l + B

2 K C l O_{3} + B 2 \to K I O_{3} + C l_{2}

A

B

{3 C l}_{2} + 6 N a O H ⟶ 5 N a C l + {N a C l O}_{3} + {3 H}_{2} O, {C l}_{2}

C l_{2} + 2 N a O H ⟶ N a C l + N a C l O + H_{2} O 3 N a C l O ⟶ 2 N a C l + N a C l O_{3} 4 N a C l O_{3} ⟶ N a C l + 3 N a C l O_{4}

N a C l O_{3}

C l_{2}

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