Question

How many milliliters of 0.230 M $\text{N}{{\text{a}}_{2}}\text{S}$ are needed to react with 30.00 mL of 0.513 M $\text{AgN}{{\text{O}}_{3}}$ , according to the following balanced equation?
\[\text{N}{{\text{a}}_{\text{2}}}\text{S(aq)}+2\text{AgN}{{\text{O}}_{\text{3}}}\text{(aq)}\to 2\text{NaN}{{\text{O}}_{\text{3}}}\text{(aq)}+\text{A}{{\text{g}}_{\text{2}}}\text{S(s)}\]

Answer 1

Hint: By using the stoichiometry of the reaction, we can find out the number of moles of $\text{N}{{\text{a}}_{2}}\text{S}$ needed to react with 30.00 mL of 0.513 M $\text{AgN}{{\text{O}}_{3}}$. The number of moles is related to volume as given below:
\[\text{Molarity (M)}=\dfrac{\text{number of moles (n)}}{\text{volume in litres (V)}}\]

Complete answer:
Stoichiometry of the chemical reactions is used to find the amount of reactant or product that is required or produced respectively with the help of their coefficient values in a balanced chemical equation.
According to the balanced chemical equation in the given question, 1 mole of $\text{N}{{\text{a}}_{2}}\text{S}$ is reacting with 2 moles of $\text{AgN}{{\text{O}}_{3}}$ to produce 2 moles of $\text{NaN}{{\text{O}}_{3}}$ and 1 mole of $\text{A}{{\text{g}}_{\text{2}}}\text{S}$.
\[\text{N}{{\text{a}}_{\text{2}}}\text{S(aq)}+2\text{AgN}{{\text{O}}_{\text{3}}}\text{(aq)}\to 2\text{NaN}{{\text{O}}_{\text{3}}}\text{(aq)}+\text{A}{{\text{g}}_{\text{2}}}\text{S(s)}\]
First, we need to find out the number of moles of $\text{N}{{\text{a}}_{2}}\text{S}$ and $\text{AgN}{{\text{O}}_{3}}$ by using this formula:
\[\text{Molarity (M)}=\dfrac{\text{number of moles (n)}}{\text{volume in litres (V)}}\]
We are given 30.00 mL of 0.513 M $\text{AgN}{{\text{O}}_{3}}$.
\[\begin{align}
  & \therefore 0.513\text{ mol }{{\text{L}}^{-1}}=\dfrac{{{\text{n}}_{\text{AgN}{{\text{O}}_{\text{3}}}}}}{0.0300\text{ L}} \\
 & \Rightarrow {{\text{n}}_{\text{AgN}{{\text{O}}_{\text{3}}}}}=0.513\text{ mol }{{\text{L}}^{-1}}\times 0.0300\text{ L} \\
 & \Rightarrow {{\text{n}}_{\text{AgN}{{\text{O}}_{\text{3}}}}}=0.01539\text{ mol} \\
\end{align}\]
Since 2 moles of $\text{AgN}{{\text{O}}_{3}}$ is reacting with 1 mole of $\text{N}{{\text{a}}_{2}}\text{S}$.
Therefore, 1 mole $\text{AgN}{{\text{O}}_{3}}$ will react with $\left( \dfrac{1}{2} \right)$ moles of $\text{N}{{\text{a}}_{2}}\text{S}$.
And, 0.01539 moles of $\text{AgN}{{\text{O}}_{3}}$ will react with $\left( \dfrac{1}{2}\times 0.01539 \right)$ i.e. 0.007695 moles of $\text{N}{{\text{a}}_{2}}\text{S}$.
So, now we need to find the volume (in mL) of 0.007695 moles of $\text{N}{{\text{a}}_{2}}\text{S}$ present in its 0.230 M solution.
\[\begin{align}
  & 0.230\text{ mol }{{\text{L}}^{-1}}=\dfrac{0.007695\text{ mol}}{{{\text{V}}_{\text{N}{{\text{a}}_{\text{2}}}\text{S}}}} \\
 & \Rightarrow {{\text{V}}_{\text{N}{{\text{a}}_{\text{2}}}\text{S}}}=\dfrac{0.007695\text{ mol}}{0.230\text{ mol }{{\text{L}}^{-1}}} \\
 & \Rightarrow {{\text{V}}_{\text{N}{{\text{a}}_{\text{2}}}\text{S}}}=0.03346\text{ L} \\
 & \Rightarrow {{\text{V}}_{\text{N}{{\text{a}}_{\text{2}}}\text{S}}}=33.46\text{ mL} \\
\end{align}\]
Hence, 33.46 milliliters of 0.230 M $\text{N}{{\text{a}}_{2}}\text{S}$ are needed to react with 30.00 mL of 0.513 M $\text{AgN}{{\text{O}}_{3}}$.

Note:
The chemical equation should always be balanced before performing the stoichiometric calculations and the number of molecules of reactant or products involved is expressed in terms of moles.