Answer
$c(NH_{4}Cl)=3.6364\frac{mol}{dm^{3}}$
Work Step by Step
$V(HCl)=21ml=0.021dm^{3}$
$c(HCl)=12\frac{mol}{dm^{3}}$
$V(NH_{3})=17.5ml=0.0175dm^{3}$
$c(NH_{3})=8\frac{mol}{dm^{3}}$---$c(NH_{4}Cl)=?$
$n(HCl)=c(HCl)×V(HCl)=12\frac{mol}{dm^{3}}×0.021dm^{3}=0.252mol$
$n(NH_{3})=c(NH_{3})×V(NH_{3})=8\frac{mol}{dm^{3}}×0.0175dm^{3}=0.14mol$
The following reaction takes place:
$HCl+NH_{3}→NH_{4}Cl$
$HCl$ and $NH_{3}$ react in 1:1 ratio. Since the amount of $NH_{3}$ is less than the amount of $HCl$, ammonia is the limiting reagent in this reaction, and therefore, the amount of ammonium chloride obtained will be equal to the initial amount of ammonia.
$n(NH_{4}Cl)=n(NH_{3})=0.14mol$
$V_{tot}=V(HCl)+V(NH_{3})=38.5ml=0.0385dm^{3}$
$\implies c(NH_{4}Cl)=\frac{n(NH_{4}Cl)}{V_{tot}}=\frac{0.14mol}{0.0385dm^{3}}\approx 3.6364\frac{mol}{dm^{3}}$
