Question

$1{\mathrm{CH}}_4+4{\mathrm{Cl}}_2\to {\mathrm{CCl}}_4+4\mathrm{HCl}$ find the equivalent mass of ${\mathrm{CH}}_4$.

Answer 1

Step 1: Given data

The following reaction is given: $1{\mathrm{CH}}_4+4{\mathrm{Cl}}_2\to {\mathrm{CCl}}_4+4\mathrm{HCl}$

Here, The reactants are methane$\left({\mathrm{CH}}_4\right)$ and chlorine$\left({\mathrm{Cl}}_2\right)$ and the products are carbon tetrachloride$\left({\mathrm{CCl}}_4\right)$ and hydrochloric acid$\left(\mathrm{HCl}\right)$.

Step 2: Formula used

For calculating the equivalent weight, the following formula is used:

$Equivalentweight=\frac{Molarmass}{nfactor}$

For calculating the number of equivalents, the following formula is used:

$no.ofEquivalents=\frac{givenweight}{equivalentweight}$

Step 3: Calculation of equivalent weight of $\mathrm{HCl}$

The $\mathrm{n}$-factor of $\mathrm{HCl}$is $1$.

The molar mass of $\mathrm{HCl}$ is $36.5$g.

Hence, applying the values in the formula for calculating the equivalent weight.

$Equivalentweight=\frac{36.5}{1}=36.5$

Step 4: Calculation of equivalent mass of ${\mathrm{CH}}_4$

Applying the formula for calculating the number of equivalents to calculate the number of equivalents of $\mathrm{HCl}$.

$\mathrm{eq}.\mathrm{of}\mathrm{HCl}=\frac{\mathit{36}\mathit{.}\mathit{5}}{\mathit{36}\mathit{.}\mathit{5}}=1$

Applying the formula for calculating the number of equivalents to calculate the number of equivalents of ${\mathrm{CH}}_4$.

$eq.ofC{H}_4=\frac{16}{eq.wt.ofC{H}_4}$

According to the law of equivalence, all reactants and products in a chemical process must have the same equivalents.

$\Rightarrow \mathrm{eq}.\mathrm{of}\mathrm{HCl}=\mathrm{eq}.\mathrm{of}{\mathrm{CH}}_4$

$$\begin{gathered} \Rightarrow \frac{16}{eq.wt.ofC{H}_4}=1 \\ \Rightarrow eq.wt.ofC{H}_4=16g \end{gathered}$$

Hence, the equivalent mass of ${\mathrm{CH}}_4$ is 16 g.