Question

For the formation of $3.65\mathrm{g}$ $\mathrm{HCl}$ gas. What volume of hydrogen gas and chlorine at $\mathrm{STP}$ are required?

Answer 1

Step 1: Given data

Mass of $\mathrm{HCl}=3.65\mathrm{g}$

Step 2: Calculating moles

Molecular mass of $\mathrm{HCl}=36.5\mathrm{g}$

Reaction to the formation of $\mathrm{HCl}$ gas $\underset{\mathrm{Hydrogen}\mathrm{gas}}{\underbrace{{\mathrm{H}}_2\left(\mathrm{g}\right)}}+\underset{\mathrm{Chlorine}\mathrm{gas}}{\underbrace{{\mathrm{Cl}}_2\left(\mathrm{g}\right)}}\to \underset{\mathrm{Hydrochloric}\mathrm{acid}}{\underbrace{2\mathrm{HCl}\left(\mathrm{l}\right)}}$

Moles of $\mathrm{HCl}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molecular}\mathrm{mass}}$

$=\frac{3.65\mathrm{g}}{36.5\mathrm{g}}=0.1\mathrm{moles}$

$\underset{\mathrm{Hydrogen}}{\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Chlorine}}{\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Hydrochloric}\mathrm{acid}}{\mathrm{HCl}\left(\mathrm{l}\right)}$

From, the reaction, $1\mathrm{mole}$ of $\mathrm{HCl}$ requires $\frac{1}{2}\mathrm{mole}$ of hydrogen and $\frac{1}{2}\mathrm{mole}$of chlorine.

Step 3: Applying stoichiometry

$0.01\mathrm{mol}$ $\mathrm{HCl}$ requires $0.05\mathrm{mol}$ hydrogen and $0.05\mathrm{mol}$ chlorine.

We know, $1\mathrm{mol}$ of gas at $\mathrm{STP}$$=$ $22.4\mathrm{L}$ of gas.

For $0.05\mathrm{mol}$ hydrogen and chlorine volume $=0.05\times 22.4=1.12\mathrm{L}$

Therefore, $1.12\mathrm{L}$ hydrogen gas and chlorine are required at $\mathrm{STP}$.