For the balanced redox reaction-2KMnO4 - 16HCl - 2KCl - 2MnCl2 - 5Cl2 - 8H2OCalculate the number of gram equivalents of HCl if 58-4 g of HCl is taken-

As only 10 moles of Cl^ ions undergoes oxidation, n factor of HCl will be = 1016 = 58 Equivalent weight of HCl = Molar Massn factor Equivalent weight of H ...

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Standard XIII

Chemistry

Equivalent Mass

For the balan...

Question

For the balanced redox reaction:

$2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 5 C l_{2} + 8 H_{2} O$
Calculate the number of gram equivalents of $H C l$ if $58.4 g$ of $H C l$ is taken.

0.25

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0.5

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0.75

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Solution

The correct option is D 1
As only 10 moles of $C l^{-}$ ions undergoes oxidation,
n - factor of HCl will be $= \frac{10}{16} = \frac{5}{8}$
$Equivalent weight of HCl = \frac{Molar Mass}{n - factor}$

$Equivalent weight of HCl = \frac{36.5}{(\frac{5}{8})} = \frac{8 \times 36.5}{5} = 58.4$
$Equivalents = \frac{Given weight}{Equivalent weight} Equivalents = \frac{58.4}{58.4} = 1$

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Equivalent Mass

Standard XIII Chemistry

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For the balanced redox reaction: 2KMnO4+16HCl→2KCl+2MnCl2+5Cl2+8H2O Calculate the number of gram equivalents of HCl if 58.4 g of HCl is taken.

The correct option is D 1As only 10 moles of Cl− ions undergoes oxidation, n - factor of HCl will be =1016=58 Equivalent weight of HCl =Molar Massn - factor Equivalent weight of HCl =36.5(58)=8×36.55=58.4 Equivalents=Given weightEquivalent weightEquivalents=58.458.4=1

For the balanced redox reaction:

$2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 5 C l_{2} + 8 H_{2} O$
Calculate the number of gram equivalents of $H C l$ if $58.4 g$ of $H C l$ is taken.