Question

How many grams of silver chloride are produced from $5.0 {\text{g}}$ of silver nitrate reacting with an excess of barium chloride in the reaction
$2AgN{O_3} + BaC{l_2} \to 2AgCl + Ba{\left( {N{O_3}} \right)_2}$?

Answer 1

Hint:This question can easily be solved with the help of the formula for finding out the number of moles. Firstly, we will find out the number of moles of silver nitrate according to the given chemical reaction. And then using stoichiometric coefficients, we will find out the mass of silver chloride.

Formula Used:
We will use the following formula to solve this question:
\[n = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}\]
Where
\[n\] is the number of moles
\[N\] is the total number of entities in the sample
\[{N_a}\] is the Avogadro’s constant

Complete step-by-step answer:Let us first write the balanced chemical equation
$2AgN{O_3}(aq) + BaC{l_2}(aq) \to 2AgCl(s) \downarrow + Ba{\left( {N{O_3}} \right)_2}(aq)$
We can observe that silver chloride is the precipitate in the given chemical equation
Let us now find out the number of moles of silver nitrate in the given reaction
By using our formula, we get
Moles of silver nitrate $ = \dfrac{{5.0 {\text{g}}}}{{169.87 {\text{g mo}}{{\text{l}}^{{\text{ - 1}}}}}} = 0.0294 {\text{mol}}$
From the chemical reaction, we can observe that the stoichiometric coefficients of silver nitrate ($AgN{O_3}$) and silver chloride ($AgCl$) are the same, that is, \[2\].
So, the moles of silver nitrate ($AgN{O_3}$) and silver chloride ($AgCl$) will also be the same, that is, $0.0294$.
Now, we can easily calculate the mass of silver nitrate by multiplying the number of moles with the molecular mass of silver nitrate.
The molar mass of silver nitrate is \[143.32 {\text{g mo}}{{\text{l}}^{{\text{ - 1}}}}\]
$0.0294{\text{ mol}} \times 143.32{\text{ g mo}}{{\text{l}}^{{\text{ - 1}}}} = 4.22 {\text{g}}$.
Hence, $4.22$ grams of silver chloride are produced.

Note:Precipitation reactions are generally double displacement reactions that involve the production of a residue of a solid form called the precipitate. These reactions also occur when two or more solutions are combined with different salts, resulting in the formation of insoluble salts from which the solution is precipitated.