Question

When a solution containing $5.85\,g$ of $NaCl$is added to a solution containing $3.4\,g$ of$AgN{O_3}$, what is the weight of $AgCl$that is precipitated $?$
(i) $2.87\,g$
(ii) $28\,g$
(iii) $58\,g$
(iv) $9.25\,g$

Answer 1

Hint:Since both of the reactants and one of the products are given try to write down a balanced chemical equation corresponding to the reaction. Next calculate the number of moles of the reactants. Check out for the limiting reactant and from that calculate the number of moles of $AgCl$precipitated and hence find the weight.

Complete step-by-step solution:The balanced chemical equation for the given reaction where $NaCl$and$AgN{O_3}$ are the reactants and $AgCl$ is one of the products which is precipitated out is given by
$NaCl\, + \,AgN{O_{3\,}}\, \to \,AgCl \downarrow \, + \,NaN{O_3}^{}...........\left( 1 \right)$
Now, we will find out the number of moles of $NaCl$and$AgN{O_3}$, and look for the limiting reagent.
Now, number moles of a compound $ = \,\dfrac{{Given\,Weight}}{{Molecular\,Weight}}$
For $NaCl$,
Given Weight $ = $ $5.85\,g$
Molecular Weight $ = $ $58.5\,g\,mo{l^{ - 1}}$
$\therefore $ number of moles of $NaCl$$ = \,\dfrac{{5.85\,g}}{{58.5\,g\,mo{l^{ - 1}}}}\, = \,0.1\,mol$
For $AgN{O_3}$,
Given Weight $ = $ $3.4\,g$
Molecular Weight $ = $ $169.87\,g\,mo{l^{ - 1}}$
$\therefore $ number of moles of $NaCl$$ = \,\dfrac{{3.4\,g}}{{169.87\,g\,mo{l^{ - 1}}}}\, = \,0.02\,mol$
Now since $0.1\,mol$ of $NaCl$ and $0.02\,mol$of $AgN{O_3}$ is present. As the number of moles of$NaCl$is more than that of$AgN{O_3}$,$NaCl$will be present in excess after the reaction is over.
Hence $AgN{O_3}$ is the limiting reagent and it will decide the number of moles of $AgCl$ that will be precipitated out.
From equation $\left( 1 \right)$we can say that $1\,mol$of $AgN{O_3}$reacts with $1\,mol$of $NaCl$ to precipitate $1\,mol$of $AgCl$ $(\,NaN{O_3}$ is not required in this case $)$.
Since $AgN{O_3}$is the limiting reagent,
Therefore, $0.02\,mol$ of $AgN{O_3}$reacts with $0.02\,mol$of $NaCl$ to precipitate $0.02\,mol$of $AgCl$.
Therefore, $0.02\,mol$of $AgCl$ precipitated out.
Now since number moles of a compound $ = \,\dfrac{{Given\,Weight}}{{Molecular\,Weight}}$
$\therefore $ Given Weight of a compound $ = $ $($Number of moles of the compound $ \times $ Molecular Weight$)\,g$
Molecular weight of $AgCl$$ = \,143.32\,g\,mo{l^{ - 1}}$
$\therefore $ Weight of $AgCl$ that is precipitated out $ = $ $\left( {0.02\,mol\, \times \,143.32\,g\,mo{l^{ - 1}}} \right)\, = \,2.87\,g$

Hence the correct answer is (i) $2.87\,g$.

Note: Write a proper balanced equation for the reaction given in the question if there is some error in the equation it will lead to error in the calculation. Finding out the limiting reagent is an important step and hence should be done properly. Also take care of the units. Try to write down the units in each and every step so that error may be avoided.