Answer:
Approximately [tex]2.53\; \rm L[/tex] (rounded to three significant figures) assuming that [tex]{\rm HCl}\, (aq)[/tex] is in excess.
Explanation:
When [tex]{\rm HCl} \, (aq)[/tex] and [tex]{\rm AgNO_3}\, (aq)[/tex] precipitate, [tex]{\rm AgCl} \, (s)[/tex] (the said precipitate) and [tex]\rm HNO_3\, (aq)[/tex] are produced:
[tex]{\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq)[/tex] (verify that this equation is indeed balanced.)
Look up the relative atomic mass of [tex]\rm Ag[/tex] and [tex]\rm Cl[/tex] on a modern periodic table:
- [tex]\rm Ag[/tex]: [tex]107.868[/tex].
- [tex]\rm Cl[/tex]: [tex]35.45[/tex].
Calculate the formula mass of the precipitate, [tex]\rm AgCl[/tex]:
[tex]\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm g\cdot mol^{-1}\end{aligned}[/tex].
Calculate the number of moles of [tex]\rm AgCl[/tex] formula units in [tex]45.00\; \rm g[/tex] of this compound:
[tex]\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}[/tex].
Notice that in the balanced equation for this reaction, the coefficients of [tex]{\rm AgNO_3} \, (aq)[/tex] and [tex]{\rm AgCl}\, (s)[/tex] are both one.
In other words, if [tex]{\rm HCl}\, (aq)[/tex] (the other reactant) is in excess, it would take exactly [tex]1\; \rm mol[/tex] of [tex]{\rm AgNO_3} \, (aq)\![/tex] formula units to produce [tex]1\; \rm mol \![/tex] of [tex]{\rm AgCl}\, (s)\![/tex] formula units.
Hence, it would take [tex]0.313987\; \rm mol[/tex] of [tex]{\rm AgNO_3} \, (aq)\![/tex] formula units to produce [tex]0.313987\; \rm mol\![/tex] of [tex]{\rm AgCl}\, (s)\![/tex] formula units.
Calculate the volume of the [tex]{\rm AgNO_3} \, (aq)\![/tex] solution given that the concentration of the solution is [tex]0.124\; \rm mol \cdot L^{-1}[/tex]:
[tex]\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}[/tex].
(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of [tex]{\rm AgNO_3} \, (aq)\![/tex].)
In other words, approximately [tex]2.53\; \rm L[/tex] of that [tex]{\rm AgNO_3} \, (aq)\![/tex] solution would be required.
