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Question: The n-factor of \({\text{HCl}}\) in the equation: \({{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{...

The n-factor of HCl{\text{HCl}} in the equation:
K2Cr2O7+14HCl2KCl+2CrCl3+3Cl2+7H2O{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}} + {\text{14HCl}} \to {\text{2KCl}} + {\text{2CrC}}{{\text{l}}_{\text{3}}} + {\text{3C}}{{\text{l}}_{\text{2}}} + {\text{7}}{{\text{H}}_{\text{2}}}{\text{O}}
A.23\dfrac{2}{3}
B.113\dfrac{{11}}{3}
C.73\dfrac{7}{3}
D.37\dfrac{3}{7}

Explanation

Solution

The n factor is also known as the valence factor and its value varies depending upon the compound being considered. For oxidizing agents and reducing agents, the n factor is equal to the number of electrons transferred by one mole of the oxidizing agent or reducing agent.

Complete step by step answer:
In the given reaction,
K2Cr2O7+14HCl2KCl+2CrCl3+3Cl2+7H2O{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}} + {\text{14HCl}} \to {\text{2KCl}} + {\text{2CrC}}{{\text{l}}_{\text{3}}} + {\text{3C}}{{\text{l}}_{\text{2}}} + {\text{7}}{{\text{H}}_{\text{2}}}{\text{O}}
Potassium dichromate is converted to chromium chloride.
In dichromate ion, chromium has an oxidation number of +6 + 6 , while in chromium chloride its oxidation number is +3 + 3 . As a result, we can say that three electrons are gained per mole of chromium atom.
Since in the given reaction, two moles of chromium atom are involved the number of electrons transferred will be =2×3=6 = 2 \times 3 = 6
Since the dichromate molecule gains 66 electrons, thus, the other reactant, that is, hydrochloric acid must lose 66 electrons.
The reaction involves 1414 moles of HCl{\text{HCl}}.
Thus, we can say that 1414 moles of HCl{\text{HCl}} lose 66electrons.
So, 11mole of HCl{\text{HCl}} loses 614\dfrac{6}{{14}}electrons.
Thus, the n factor of HCl{\text{HCl}} is 37\dfrac{3}{7}.

The correct answer is D.
Note:
In general, chemicals combine in definite ratios in the chemical reactions. Since chemical reactions can neither create nor destroy matter, nor transmute one element into another, thus the amount of each element must be the same throughout the entire reaction. For instance, the number of atoms of a given element X on the reactant side must be equal to the number of atoms of that element on the product side, irrespective of whether or not all of those atoms are involved in a reaction.