Question

For the balanced redox reaction:

$$ 2 \text{KMnO}_4 + 16 \text{HCl} \rightarrow 2 \text{KCl} + 2 \text{MnCl}_2 + 5 \text{Cl}_2 + 8 \text{H}_2 \text{O} $$

Calculate the number of gram equivalents of (\text{HCl}) if 58.4 g of (\text{HCl}) is taken. A. 0.25 B. 0.5 C. 0.75 D. 1

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Answer

The correct answer is D. 1.

To solve this, let's go through the steps:

  1. Identify change in oxidation state: Recognize that only 10 moles of $\mathrm{Cl}^-$ ions undergo oxidation, and there is no change in the oxidation state ($ \mathrm{O.S.} $).

  2. Determine the n-factor of $\mathrm{HCl}$: $$ n\text{-factor of }\mathrm{HCl} = \frac{10 \text{ moles}}{16 \text{ moles}} = \frac{5}{8} $$

  3. Calculate the equivalent weight of $\mathrm{HCl}$: $$ \text{Equivalent weight of }\mathrm{HCl} = \frac{\text{Molar Mass}}{n\text{-factor}} $$ Given that the molar mass of $\mathrm{HCl}$ is 36.5 g/mol, $$ \text{Equivalent weight of }\mathrm{HCl} = \frac{36.5}{\left(\frac{5}{8}\right)} = \frac{8 \times 36.5}{5} = 58.4 \text{ g/equiv} $$

  4. Calculate the number of gram equivalents: $$ \text{Gram equivalents} = \frac{\text{Given weight}}{\text{Equivalent weight}} $$ Given that the weight of $\mathrm{HCl}$ is 58.4 g, $$ \text{Gram equivalents} = \frac{58.4 \text{ g}}{58.4 \text{ g/equiv}} = 1 $$

Therefore, the number of gram equivalents of $\mathrm{HCl}$ in 58.4 g of $\mathrm{HCl}$ is 1.


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