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:

(a) The mass of calcium chloride formed

(b) The volume, at STP of ${\mathrm{NH}}_3$ liberated.

Answer 1

Step 1 Calculating the mass of Calcium chloride formed:

$\underset{\mathrm{Calcium}\mathrm{hydroxide}}{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2}+\underset{\mathrm{Ammonium}\mathrm{chloride}}{2{\mathrm{NH}}_4\mathrm{Cl}}\to \underset{\mathrm{Calcium}\mathrm{chloride}}{{\mathrm{CaCl}}_2}+\underset{\mathrm{Ammonia}}{2{\mathrm{NH}}_3}+\underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}}$

$$\begin{gathered} \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{NH}}_4\mathrm{Cl}=53.5\mathrm{g}{\mathrm{mol}}^{-1} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{CaCl}}_2=111\mathrm{g}{\mathrm{mol}}^{-1} \\ \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{NH}}_4\mathrm{Cl}=\frac{5.35\mathrm{g}}{53.5\mathrm{g}{\mathrm{mol}}^{-1}}=0.1\mathrm{mole} \end{gathered}$$

From the given equation,

2 moles ${\mathrm{NH}}_4\mathrm{Cl}$ give 1 mole ${\mathrm{CaCl}}_2$

0.1 moles of ${\mathrm{NH}}_4\mathrm{Cl}$ will react to give $\frac{1}{2}\times 0.1=0.05\mathrm{moles}$ ${\mathrm{CaCl}}_2$

$$\begin{gathered} \mathrm{Mass}\mathrm{of}{\mathrm{CaCl}}_2=\mathrm{Number}\mathrm{of}\mathrm{moles}\times \mathrm{Molar}\mathrm{mass} \\ \mathrm{Mass}\mathrm{of}{\mathrm{CaCl}}_2=0.05\mathrm{mol}\times 111\mathrm{g}{\mathrm{mol}}^{-1} \\ \mathrm{Mass}\mathrm{of}{\mathrm{CaCl}}_2=5.55\mathrm{g} \end{gathered}$$

Step 2: Calculating the volume of ${\mathrm{NH}}_3$ liberated at STP

From the given equation,

$2\mathrm{moles}$ of ${\mathrm{NH}}_4\mathrm{Cl}$ will react to give $2\mathrm{moles}$ ${\mathrm{NH}}_3$

So, $0.1\mathrm{moles}$ of ${\mathrm{NH}}_4\mathrm{Cl}$ will react to give $0.1\mathrm{moles}$${\mathrm{NH}}_3$

At STP,

The volume of one mole ${\mathrm{NH}}_3$$=22.4\mathrm{L}$

Volume of $0.1$ mole ${\mathrm{NH}}_3$ $=22.4\times 0.1=2.24\mathrm{L}$ ${\mathrm{NH}}_3$

Hence,

The mass of calcium chloride formed $5.55\mathrm{g}$

The volume, at STP of ${\mathrm{NH}}_3$ liberated $2.24\mathrm{L}$.