What is the equivalent mass of HCl in the given reaction?
$2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 5 C l_{2} + 8 H_{2} O$

44.2

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36.5

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58.4

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none of these

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Solution

The correct option is C 58.4
As only 10 moles of $C l^{-}$ ions undergoes oxidation,
n - factor of HCl will be $= \frac{10}{16} = \frac{5}{8}$
$Equivalent weight of HCl = \frac{Molar Mass}{n - factor}$
$Equivalent weight of HCl = \frac{36.5}{(\frac{5}{8})} = \frac{8 \times 36.5}{5} = 58.4$

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Similar questions

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 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

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

H C l

58.4 g

H C l

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

H C l

58.4 g

H C l

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

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