Question

When an excess of sodium hydroxide is added to an aqueous solution of ammonium chloride, gaseous ammonia is produced :

$\mathbf{NaOH}\left(\mathbf{aq}\right)\mathbf{+}{\mathbf{NH}}_{\mathbf{4}}\mathbf{Cl}\left(\mathbf{aq}\right)\to \mathbf{NaCl}\left(\mathbf{aq}\right)\mathbf{+}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right)$

Suppose 3.68 g ammonium chloride reacts in this way at ${\mathbf{30}}^{\mathbf{o}}\mathbf{C}$and a total pressure of 0.9984 atm . At this temperature, the vapor pressure of water is 0.0419 atm. Calculate the volume of ammonia saturated with water vapor that will be produced under these conditions, assuming no leaks or other losses of gas.

Short answer

The volume of ammonia is 1.84 L.

Step-by-step solution

Given data

Total pressure = 0.9884 atm

Pressure due to ${\text{H}}_2\text{O vapour}=0.0419 \text{atm}$

Mass of$\text{NH}4\text{Cl}=3.68\text{g}$

Temperature$=\text{T}=30°\text{C}=(30+273.15)\text{K}=303.15\text{K}$

Concept of Ideal gas equation

The ideal gas law shows the relation between various parameters of the gas, pressure, volume, moles and temperature. The mathematical expression of ideal gas is: PV = nRT , Where

$$\begin{gathered} P=\mathrm{Pressure} \\ V=\mathrm{volume}\left(\mathrm{L}\right) \\ T=\mathrm{temperature} \\ R=0.08206 \mathrm{L}.\mathrm{atm}/\mathrm{K}.\mathrm{mol} \end{gathered}$$

Calculation of number of moles of ammonium chloride

Molar mass of${\text{NH}}_4\text{Cl}=52.5\text{g}/{\text{mol}}_{}$${\text{NH}}_4\text{Cl}=52.5\text{g}/{\text{mol}}_{}$

No. of moles of${\text{NH}}_4\text{Cl}=\frac{3.68}{52.5}=0.07\text{mol}$

So, number of moles of ${\text{NH}}_4\text{Cl=0.07mol}$

Calculation of volume of ammonia

So, from the balanced equation between ammonium chloride and sodium hydroxide We see that, 1 mole of${\text{NH}}_3\text{(g)}$ gas is produced by 1 mole of ${\text{NH}}_4\text{Cl}$.

$\text{NaOH}\left(aq\right)+{\text{NH}}_4\text{Cl}\left(\text{aq}\right)\to \text{NaCl}\left(aq\right)+{\text{NH}}_3\left(g\right)+{\text{H}}_2\text{O}\left(l\right)$

So, by using Ideal gas equation, we calculate volume of ammonia

$R=0.08206 \text{Latm}/\text{Kmol}$

$\begin{array}{rcl}V& =& \frac{nRT}{P}\\ & =& \frac{\text{0.07mol}\times \text{0.08206} \text{Latm/Kmol}\times \text{303.15} \text{K}}{\text{0.9465} \text{atm}}\\ V& =& 1.84\text{L}.\end{array}$

The volume of ammonia saturated with water vapour is $\begin{array}{rcl}& & 1.84\text{L}\end{array}$