Question

When ammonia reacts with hydrogen chloride gas, it produces white fumes of ammonium chloride. The volume occupied by $N{H}_3$ in glass bulb $A$ is three times more than the volume occupied by $HCl$ in glass bulb $B$ at STP.
i) How many moles of ammonia are present in glass bulb $A$?
ii) How many grams of $N{H}_{3}Cl$ will be formed when the stopper is opened?
(Atomic mass of $N=14,H=1,Cl=35.5)$
iii) Which gas will remain after completion of the reaction?
iv) Write the chemical reaction involved in this process.

Answer 1

(i) $1$ mole occupies $22.4L$ volume

$\therefore$ moles of $N{H}_3$ present= $\frac{67.2}{22.4}=3$

(ii) $1$ mole of $N{H}_3$ reacts with $1$ mole of $HCl$ to give $1$ mole of $N{H}_{4}Cl$. Here only $1$ mole of $HCl$ & $3$ moles of $N{H}_3$ are present. Hence, $1$ mole of $N{H}_{4}Cl$ is produced.

Mass of $N{H}_{4}Cl$ produced= $1\times$ Molecular mass

$=1\times (14+4\times 1+35.5)$

$=53.5g$

(iii) Here $N{H}_3$ is present in excess than what is required to react with $1$ mole $HCl$. Hence $3-1=2$ moles of $N{H}_3$ will remain after completion of the reaction.

(iv) Chemical reaction involved is:

$N{H}_3+HCl⟶N{H}_{4}Cl$ .