Question

When $22.4$ litres of ${\mathrm{H}}_2\left(\mathrm{g}\right)$ is mixed with $11.2$ litres of ${\mathrm{Cl}}_2\left(\mathrm{g}\right)$, each at $\mathrm{STP}$, the moles of $\mathrm{HCl}\left(\mathrm{g}\right)$ formed are equal to

• A. $1\mathrm{mol}\mathrm{of}\mathrm{HCl}\left(\mathrm{g}\right)$
• B. $2\mathrm{mol}\mathrm{of}\mathrm{HCl}\left(\mathrm{g}\right)$
• C. $0.5\mathrm{mol}\mathrm{of}\mathrm{HCl}\left(\mathrm{g}\right)$
• D. $1.5\mathrm{mol}\mathrm{of}\mathrm{HCl}\left(\mathrm{g}\right)$

Answer 1

The correct option is A

$1\mathrm{mol}\mathrm{of}\mathrm{HCl}\left(\mathrm{g}\right)$

Explanation for correct option:

(A) $1\mathrm{mol}\mathrm{of}\mathrm{HCl}\left(\mathrm{g}\right)$

The Avogadro's law for the equation:

• According to Avogadro's law, equal volumes of all gases contain equal numbers of molecules under equivalent conditions. Because one mole of molecules in all gases has $6.022\mathrm{x}{10}^{23}$molecules, at the same pressure and temperature, they all occupy the same volume.
• At STP, one mole of any gas occupies a volume of $22.4\mathrm{L}$.

${\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to 2\mathrm{HCl}\left(\mathrm{g}\right)$

Limiting reagent in the reaction:

• A limiting reagent in a reaction is a species that is present in less than the amount required by the reaction's stoichiometric relation. The reaction's limiting reagent is chlorine.

Mole calculation for $\mathrm{HCl}\left(\mathrm{g}\right)$:

• In this reaction, chlorine is the limiting reagent as it contains less number of $11.2\mathrm{L}$ moles than compared to hydrogen.

$$\begin{gathered} (\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{H}}_2){\mathrm{nH}}_2=\frac{22.4}{22.4};(\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{Cl}}_2){\mathrm{nCl}}_2=\frac{11.2}{22.4} \\ {\mathrm{nH}}_2=1\mathrm{mol}{\mathrm{nCl}}_2=0.5\mathrm{mol} \\ {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to 2\mathrm{HCl}\left(\mathrm{g}\right) \\ \mathrm{Initial} \\ (\mathrm{Before}1\mathrm{mol}0.5\mathrm{mol}0\mathrm{mol} \\ \mathrm{Reaction}) \\ \mathrm{Final} \\ (\mathrm{After}(1-0.5)(0.5-0.5)2\mathrm{x}0.5 \\ \mathrm{reaction})=0.5\mathrm{mol}=0\mathrm{mol}=1\mathrm{mol} \end{gathered}$$

Explanation for incorrect option:

The moles of HCl so formed is $1\mathrm{mol}$. Thus options B, C, and D are incorrect.

Hence, the correct option is (A) $1\mathrm{mol}\mathrm{of}\mathrm{HCl}\left(\mathrm{g}\right)$.