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 weight of calcium chloride is formed.

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

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:

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

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

Step 2: Calculation of mass of ammonium chloride and calcium chloride

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

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}1\mathrm{mol}{\mathrm{CaCl}}_2& =& 40\mathrm{g}+35.5\mathrm{g}\times 2\\ & =& 111\mathrm{g}\end{array}$

Step 3: Calculation of required weight of calcium chloride from stoichiometric calculation

• $107\mathrm{g}$ of ammonium chloride produces $111\mathrm{g}$ of calcium chloride.
• Thus, $5.35\mathrm{g}$ of ammonium chloride will produce $\frac{111}{107}\times 5.35=5.55\mathrm{g}$ of calcium chloride.

Therefore, $5.55\mathrm{g}$ of calcium chloride will be formed.