Question

Show that the reaction of ammonia and hydrochloric acid to form ammonium chloride is in accordance with the law of conservation of mass. $(\mathrm{N}=14,\mathrm{Cl}=35.5,\mathrm{H}=1)$
${\mathrm{NH}}_3+\mathrm{HCl}\rightleftharpoons {\mathrm{NH}}_4\mathrm{Cl}$

Answer 1

Law of conservation of mass:

1. It was given by Antoine Laurent Lavoisier in 1789.
2. According to him during the chemical reaction in an isolated system mass can neither be created nor can be destroyed it gets transformed from one form of matter to another.
3. Mass of reactants = Mass of products
4. Reaction:

$\underset{\mathrm{Ammonia}}{{\mathrm{NH}}_3\left(\mathrm{g}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\mathrm{HCl}\left(\mathrm{aq}\right)}\rightleftharpoons \underset{\mathrm{Ammonium}\mathrm{chloride}}{{\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right)}$

Given;$(\mathrm{N}=14\mathrm{amu},\mathrm{Cl}=35.5\mathrm{amu},\mathrm{H}=1\mathrm{amu})$

Step 1. Calculate the mass of all the compounds involved in the reaction.

i. Mass of ${\mathrm{NH}}_3$ $=$Mass of $\mathrm{N}$ $+$ $3\mathrm{x}$ mass of $\mathrm{H}$

$=$$14$ $+$ $3\mathrm{x}1$

$=$$17\mathrm{amu}$

ii. Mass of $\mathrm{HCl}$ $=$Mass of H $+$ mass of $\mathrm{Cl}$

$=$$1$ $+$ $35.5$

$=$$36.5\mathrm{amu}$

iii. Mass of ${\mathrm{NH}}_4\mathrm{Cl}$ $=$Mass of N $+$ $4\mathrm{x}$ mass of the $\mathrm{H}$ $+$ mass of $\mathrm{Cl}$

$=$$14$ $+$ $4\mathrm{x}1$ $+$ $35.5$

$=$$53.5\mathrm{amu}$

Step 2. Equate the mass of reactants and products.

Mass of reactants $=$Mass of products

Mass of ${\mathrm{NH}}_3$ $+$Mass of $\mathrm{HCl}$ $=$Mass of ${\mathrm{NH}}_4\mathrm{Cl}$

$17\mathrm{amu}$ $+$ $36.5\mathrm{amu}$ $=$$53.5\mathrm{amu}$

$53.5\mathrm{amu}$ $=$$53.5\mathrm{amu}$

Thus, the reaction of ammonia and hydrochloric acid to form ammonium chloride is by the law of conservation of mass.