Question

Calcium hydroxide reacts with ammonium chloride to give ammonia, according to the following equation:

$\mathrm{Ca}{\left(\mathrm{OH}\right)}_2+2{\mathrm{NH}}_4\mathrm{Cl}\to {\mathrm{CaCl}}_2+2{\mathrm{NH}}_3+2{\mathrm{H}}_2\mathrm{O}$

If $5.35\mathrm{g}$ of ammonium chloride is used, calculate the volume at STP of ammonia liberated.

$\left[\mathrm{H}=1,\mathrm{N}=14,\mathrm{O}=16,\mathrm{Cl}=35.5,\mathrm{Ca}=40\right]$

Answer 1

Step 1: Given information

• Calcium hydroxide $\left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\right)$ reacts with ammonium chloride $\left({\mathrm{NH}}_4\mathrm{Cl}\right)$ to give ammonia $\left({\mathrm{NH}}_3\right)$, calcium chloride $\left({\mathrm{CaCl}}_2\right)$ and water $\left({\mathrm{H}}_2\mathrm{O}\right)$.
• The given reaction is as follows:

$\mathrm{Ca}{\left(\mathrm{OH}\right)}_2+2{\mathrm{NH}}_4\mathrm{Cl}\to {\mathrm{CaCl}}_2+2{\mathrm{NH}}_3+2{\mathrm{H}}_2\mathrm{O}$

• Mass of Ammonium chloride $\left({\mathrm{NH}}_4\mathrm{Cl}\right)=5.35\mathrm{g}$

Step 2: Calculation of mass of ammonium chloride

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}2\mathrm{mol}{\mathrm{NH}}_4\mathrm{Cl}& =& 2\times \left(14\mathrm{g}+1\mathrm{g}\times 4+35.5\mathrm{g}\right)\\ & =& 107\mathrm{g}\end{array}$

Step 3: STP

• STP indicates Standard Temperature and Pressure.
• At STP, temperature$=0^{°}\text{C}$
• At STP, pressure$=1\text{atm}$
• The volume of a gas at STP is $22.4\text{L}$ or $22.4{\text{dm}}^3$.

Step 4: Calculation of volume of liberated ammonia

• From the chemical reaction, it is clear that $2$ moles of ammonium chloride liberate $2$ moles of ammonia.
• $107\mathrm{g}$ of ammonium chloride liberates $22.4\mathrm{L}\times 2=44.8\mathrm{L}$ of ammonia at STP (Standard temperature and pressure).
• Thus, $5.35\mathrm{g}$ of ammonium chloride will liberate $\frac{44.8\mathrm{L}}{107\mathrm{g}}\times 5.35\mathrm{g}=2.24\mathrm{L}$ of ammonia at STP.

Therefore, $2.24\mathrm{L}$ of ammonia will liberate at STP.