Question

A $5.0L$flask contains a mixture of ammonia and nitrogen at $27°c$ and a total pressure of role="math" localid="1663409403097" $3.00atm$ . The sample of gas is allowed to flow from the flask until the pressure in the flask has fallen to role="math" localid="1663409415998" $1.00atm$ . The gas that escapes is passed through role="math" localid="1663409428285" $1.50L$of role="math" localid="1663409439762" $0.200$acetic acid. All the ammonia in the gas that escapes is absorbed by the solution and turns out to be just sufficient to neutralize the acetic acid present. The volume of the solution does not change significantly.

(a) Will the electrical conductivity of the aqueous solution change significantly as the gas is absorbed? Give equations for any reactions, and calculate the final concentrations of the principal ions present (if any) at the end.

(b) Calculate the percentage by mass of ammonia in the flask initially.

Short answer

a) The balanced chemical equation is as follows;

${\text{NH}}_3+{\text{CH}}_3\text{COOH}\to {\text{CH}}_3{\text{COO}}^{-}+{\text{NH}}_4^{+}$

Amount of ammonia is $1.87g$ .

b) Percentage of ammonia is 73.12%.

Step-by-step solution

Given data

The volume of ammonia is $5.0L$, the total pressure is $3.00\mathrm{tm}$, the flask pressure fallen to $3.00\mathrm{tm}$, the gas that escapes is passed through $1.50L$ of $0.200\text{M}$ acetic acid.

Concept used

The moles of ammonia are calculated before and after the reaction, by using the ideal gas equation and the amount of ammonia escaped from the reaction is calculated.

The number of moles of ammonia that have escaped is the amount necessary to neutralize the acetic acid. The final concentration was calculated by using the obtained values.

Calculate the initial mole of ammonia

a)

The balanced chemical equation is as follows;

${\text{NH}}_3+{\text{CH}}_3\text{COOH}\to {\text{CH}}_3{\text{COO}}^{-}+{\text{NH}}_4^{+}$

The salt dissociated in water gives ions and conducts electricity.

Using ideal gas equation,

$$\begin{gathered} n=\frac{PV}{RT} \\ T=27+273=300\text{K} \\ P=3.0\text{atm} \\ V=5.0\text{L} \\ R=0.0821{\text{Latmmol}}^{-1}{K}^{-1} \end{gathered}$$

The values are substituted.

data-custom-editor="chemistry" $$\begin{gathered} n=\frac{3.0\text{atm}\times 5.0\text{L}}{0.0821{\text{Latmmol}}^{-1}{K}^{-1}\times 300K} \\ =0.61\text{mol} \end{gathered}$$

So, the initial moles of ${\text{NH}}_3=0.61\text{mol}$.

Calculate the moles of ammonia after reaction

Number of moles of ammonia after the reaction is found by using ideal gas equation.

$$\begin{gathered} V=5L \\ P=1.0\text{atm} \\ T=273+27=300\text{K} \end{gathered}$$

Using ideal gas equation:

$$\begin{gathered} n=\frac{PV}{RT} \\ =\frac{1.0\text{atm}\times 5.0\text{L}}{0.0821{\text{LatmmalK}}^{-1}\times 300\text{K}} \\ =0.20\text{mol} \end{gathered}$$

Moles of ammonia that escapes,

$n=0.61-0.20=0.41mol$

Calculate the moles of acetic acid

The number of moles of ammonia which have escaped is the amount necessary to neutralize the acetic acid.

Amount of acetic acid is 1.50L with molarity 0.20 mol $L^{-1}$.

Moles of acetic acid $\text{=1.50}\hspace{0.33em}\text{L×0.2mol/L=0.3mol}$

So, the composition of the gas escaping is as follows;

${\text{x}}_{\text{a}}\text{=}\frac{\text{0.30mol}}{\text{0.41mol}}\text{= 0.73}$

Calculate the amount of ammonia

Molar mass of ${\text{NH}}_3=17.03\text{g}{\text{mol}}^{-1}$

Amount of ammonia =

Number of ammonia unreacted $=0.41-0.31$$=0.1\mathrm{mol}$

So, the amount of ammonia $=0.11\text{mol}\times 17.03\text{g}{\text{mol}}^{-1}=1.87\text{g.}$

Therefore, amount of ammonia is $1.87g$.

Calculate the percentage of ammonia

(b)

Total amount of ammonia $=5.11\text{g}+1.87\text{g}=6.98\text{g}$

So, the reacted percentage of ammonia as follows;

role="math" localid="1663410498332" $$\begin{gathered} {\text{Percentage of NH}}_3=\frac{5.11\text{g}}{6.988}\times 100 \\ =73.12\% \end{gathered}$$