Question

$50 mL$ of ammonia was decomposed by a series of electric sparks and the volume after decomposition was $97 mL$. On washing the gas with water and drying the volume became $94 mL$.

To this gas was added $100 mL$ of oxygen and the mixture exploded. After allowing the gas to cool to room temperature the volume was $88.25 mL$. Derive the formula of ammonia.

Answer 1

SOLUTION : [In this problem, it is necessary to know the volume of ammonia obtained when a given volume of each of $N _{2}$ and $H _{2}$ reacts.]

We know that ammonia on decomposition is converted to $N _{2}$ and $H _{2}$ and ammonia is soluble in water, $N _{2}$ and $H _{2}$ are not.

Vol. of ammonia $=50 mL ; \quad$ Vol. after decomposition $=97 mL$

Vol. after treatment with water $=94 mL$

$\therefore$ Volume of ammonia present in $97 mL =(97-94) mL =3 mL$

This is, $3 mL$ of ammonia remained undecomposed. Hence the volume of ammonia that decomposed to give $94 mL$ of a mixture of $N_{2}$ and $H_{2}=(50-3) mL =47 mL$.

Vol. of $O_{2}$ added $=100 mL$.

$\therefore$ Total vol. containing $N_{2}, H_{2}$ and $O_{2}=(100+94) mL =194 mL$

This mixture is exploded. Only $H _{2}$ and $O _{2}$ will react to form water. iن2 will remain unreacted. Duc to formation of water, there will be contraction in volume.

Vol. after explosion $\quad=88.25 mL$

$\therefore$ Contraction in volume $\quad=(194-88.25) mL =105.75 mL$

$\therefore$ Vol. of $H _{2}$ required for this contraction $=\frac{2}{3} \times 105.75 mL =70.5 mL$

and vol. of $O _{2}$ in water formation $=\frac{1}{3} \times 105 \cdot 75 mL =35 \cdot 25 mL$.

Vol. of $O _{2}$ taken $=100 mL$ and only $35.25 mL$ reacted. It means that all the $H _{2}$ present in the mixture has completely reacted and unreacted oxygen $=(100-35 \cdot 25) mL$ $=64.75 mL$ and it must be present in the final volume, $88.25 mL$

Vol. of $N _{2}$ formed as a result of decomposition of $47 mL$ of ammonia $=(88 \cdot 25-$ $64 \cdot 75) mL =23 \cdot 50 mL$

Since $70 \cdot 5 mL$ of $H _{2}$ been obtained by decomposition of ammonia (and not from any other source), we conclude that

$23.5 mL N _{2}$ and $70.5 mL H _{2}$ combine to give $47 mL$ ammonia.

or, $\quad 1 mL N _{2}+3 mL H _{2}=2 mL$ ammonia or, $\quad 1$ molecule of $N _{2}+3$ molecules of $\quad H _{2}=2$ molecules of ammonia or, $\quad 2$ atoms of $N _{2}+6$ atoms of $\quad H _{2}=2$ molecules of ammonia. Hence, one molecule of ammonia contains one atom of nitrogen and 3 atoms of hydrogen.

Hence, one molecule of ammonia conta gen. The formula of ammonia $= NH _{3}$.

The formula of ammonia $= NH _{3}$.