In the following reaction- 2KMnO4-16HCl- 5Cl2-2MnCl2-2KCl-8H2O- the reduction product is -

The correct option is C MnCl_2 KMn^+7O_4→ Mn^+2Cl_2 .Therefore, MnCl_2 is the reduction product. ...

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Question

In the following reaction:
$2 K M n O_{4} + 16 H C l \to 5 C l_{2} + 2 M n C l_{2} + 2 K C l + 8 H_{2} O$ , the reduction product is :

C l_{2}

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M n C l_{2}

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H_{2} O

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K C l

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2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 5 C l_{2} + 8 H_{2} O

K M n O_{4} + H_{2} S O_{4} + H C l \to K_{2} S O_{4} + M n S O_{4} + C l_{2} + H_{2} O

2 K M n O_{4} + 10 K C l + 8 H_{2} S O_{4} \to 2 M n S O_{4} + 6 K_{2} S O_{4} + 8 H_{2} O + 5 C l_{2}

What is the equivalent mass of HCl in the given reaction?

$2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 5 C l_{2} + 8 H_{2} O$

{2 K}_{2} M n O_{4} + {C l}_{2} ⟶ 2 K C l + 2 K M n O_{4}

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