Question

If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L

HCL(aq) and 0.124 mol/L AgNO3(aq) what volume in litres of AgNO3 did we
add?

Answer 1

Approximately
`2.53\; \rm L` (rounded to three significant figures) assuming that
`{\rm HCl}\, (aq)` is in excess.

Step-by-step explanation:

When
`{\rm HCl} \, (aq)` and
`{\rm AgNO_3}\, (aq)` precipitate,
`{\rm AgCl} \, (s)` (the said precipitate) and
`\rm HNO_3\, (aq)` are produced:

`{\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq)` (verify that this equation is indeed balanced.)

Look up the relative atomic mass of
`\rm Ag` and
`\rm Cl` on a modern periodic table:

• 
  `\rm Ag`:
  `107.868`.
• 
  `\rm Cl`:
  `35.45`.

Calculate the formula mass of the precipitate,
`\rm AgCl`:

`\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^(-1) \\\ &\approx 143.318 \; \rm g\cdot mol^(-1)\end{aligned}`.

Calculate the number of moles of
`\rm AgCl` formula units in
`45.00\; \rm g` of this compound:

`\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx (45.00\; \rm g)/(143.318\; \rm g \cdot mol^(-1))\approx 0.313987\; \rm mol \end{aligned}`.

Notice that in the balanced equation for this reaction, the coefficients of
`{\rm AgNO_3} \, (aq)` and
`{\rm AgCl}\, (s)` are both one.

In other words, if
`{\rm HCl}\, (aq)` (the other reactant) is in excess, it would take exactly
`1\; \rm mol` of
`{\rm AgNO_3} \, (aq)\!` formula units to produce
`1\; \rm mol \!` of
`{\rm AgCl}\, (s)\!` formula units.

Hence, it would take
`0.313987\; \rm mol` of
`{\rm AgNO_3} \, (aq)\!` formula units to produce
`0.313987\; \rm mol\!` of
`{\rm AgCl}\, (s)\!` formula units.

Calculate the volume of the
`{\rm AgNO_3} \, (aq)\!` solution given that the concentration of the solution is
`0.124\; \rm mol \cdot L^(-1)`:

`\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx (0.313987\; \rm mol)/(0.124\; \rm mol \cdot L^(-1))\approx 2.53\; \rm L\end{aligned}`.

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of
`{\rm AgNO_3} \, (aq)\!`.)

In other words, approximately
`2.53\; \rm L` of that
`{\rm AgNO_3} \, (aq)\!` solution would be required.