Question

If the equivalent mass of HCl in the reaction ${K_2}C{r_2}{O_7} + HCl \to KCl + CrC{l_3} + C{l_2} + {H_2}O$ is $\dfrac{M}{x}$ where M is the molar mass of $HCl$. What is the value of x.

Answer 1

Hint: We have to know that x represents the equivalent weight. We can calculate equivalent weight using n-factor and molar mass of hydrochloric acid. Here n-factor is nothing but the total number of change in oxidation number and total number of moles of hydrochloric acid. When we divide the total number of change in oxidation number by the total number of moles of hydrochloric acid we obtain the n-factor.

Complete step by step answer:
When potassium dichromate is reacted with hydrochloric acid potassium chloride, chromium (III) chloride, chlorine and water are formed as products. We can give the chemical reaction is,
${K_2}C{r_2}{O_7} + HCl \to KCl + CrC{l_3} + C{l_2} + {H_2}O$
We can write the balanced reaction is,
${K_2}C{r_2}{O_7} + 14HCl \to 2KCl + 3CrC{l_3} + 3C{l_2} + 7{H_2}O$
In this reaction, we can see that hydrochloric acid is oxidized to chlorine.
Change in oxidation number= $0 - \left( { - 1} \right) = + 1$ . There is a change in oxidation state from hydrochloric acid to chlorine that is $ + 1$.
The number of atoms oxidized/reduced in the reaction is 6. We can also say the total number of changes in oxidation state is six.
We know that the mass of one mole of hydrochloric acid is 36.5g/mol.
The total number of moles of hydrochloric acid is 14.
We can write the n-factor for HCl as $\dfrac{6}{{14}}$.
The n-factor is the number of moles of Hydrochloric acid and total number of changes in oxidation.
Let us now calculate the equivalent mass of hydrochloric acid. The product of the number of moles of HCl present in the reaction and the mass of one mole of hydrochloric acid divided by the total number of change in oxidation gives the equivalent mass of hydrochloric acid.
Equivalent mass of hydrochloric acid =
$\dfrac{{{\text{Number of moles of HCl}} \times {\text{Mass of one mole of HCl}}}}{{{\text{Total number of change in oxidation}}}}$
Equivalent mass of hydrochloric acid = $\dfrac{{14 \times 36.5}}{6} = 85.167$
Equivalent mass of hydrochloric acid = $85.1$
The equivalent mass of hydrochloric acid is $85.1$.

Note: For elements that have more than one valency in forming compounds contains more than one equivalent weight.
For example, if iron has a valence of three, then equivalent weight would be \[18.615g/eq\]. If iron has a valence of two, then equivalent weight would be \[27.9225g/eq\].
We are calculating the values of equivalent weight with the help of the atomic mass of iron \[\left( {55.854g} \right)\] and the number of electrons shared.
The equivalent weight is used to determine the weight proportion in which is combined to form compounds. We can use equivalent weight in volumetric and gravimetric analysis. We can also use it in polymer chemistry.