Answer:
Approximately [tex]0.121\; \rm mol[/tex].
Explanation:
Balance the equation
[tex]\rm Cu\; (s)[/tex] reacts with [tex]\rm AgNO_3\; (aq)[/tex] to produce [tex]\rm Ag\; (s)[/tex] and [tex]\rm Cu(NO_3)_2\; (aq)[/tex].
[tex]\rm ? \; Cu\, (s) + ?\; AgNO_3\, (aq) \to ?\; Ag\, (s) + ?\; Cu(NO_3)_2\, (aq)[/tex].
To balance this equation, start by setting the coefficient of the most complex species to [tex]1[/tex]. For example, [tex]\rm Cu(NO_3)_2\, (aq)[/tex] has more atoms in each of its formula unit than any other species in this reaction. Let the coefficient of
[tex]\rm ? \; Cu\, (s) + ?\; AgNO_3\, (aq) \to ?\; Ag\, (s) + 1\; Cu(NO_3)_2\, (aq)[/tex].
That formula unit of [tex]\rm Cu(NO_3)_2\, (aq)[/tex] would include:
- [tex]1 \times 1 = 1[/tex] [tex]\rm Cu[/tex] atom,
- [tex]1 \times 2= 2[/tex] [tex]\rm N[/tex] atoms, and
- [tex]1 \times 2 \times 3 = 6[/tex] [tex]\rm O[/tex] atoms.
The other product, [tex]\rm Ag\; (s)[/tex], contains neither [tex]\rm Cu[/tex] atoms nor [tex]\rm N[/tex] atoms. Therefore, the product side would include exactly
- one [tex]\rm Cu[/tex] atom,
- two [tex]\rm N[/tex] atoms, and
- six [tex]\rm O[/tex] atoms.
Atoms are conserved in a chemical reaction. The reactant side shall also include:
- [tex]1 \times 1 = 1[/tex] [tex]\rm Cu[/tex] atom,
- [tex]1 \times 2= 2[/tex] [tex]\rm N[/tex] atoms, and
- [tex]1 \times 2 \times 3 = 6[/tex] [tex]\rm O[/tex] atoms.
Among the reactants, [tex]\rm AgNO_3\, (aq)[/tex] is the only source of [tex]\rm N[/tex] atoms. Each formula unit of
Similarly, the coefficient of [tex]\rm Cu[/tex] would be [tex]1/ 1 = 1[/tex].
[tex]\rm 1 \; Cu\, (s) + 1\; AgNO_3\, (aq) \to ?\; Ag\, (s) + 1\; Cu(NO_3)_2\, (aq)[/tex].
There would be exactly [tex]2[/tex] [tex]\rm Ag[/tex] atom on the reactant side (from [tex]\rm AgNO_3\, (aq)[/tex].) The products should also include
Hence the equation:
[tex]\rm 1 \; Cu\, (s) + 1\; AgNO_3\, (aq) \to 2\; Ag\, (s) + 1\; Cu(NO_3)_2\, (aq)[/tex].
Calculate n(Cu(NO₃)₂ (aq))
The ratio between the coefficients of [tex]\rm Cu(NO_3)_2\; (aq)[/tex] and [tex]\rm Ag\, (s)[/tex] is:
[tex]\displaystyle \frac{n(\mathrm{Cu(NO_3)_2\, (aq)})}{n(\mathrm{Ag\, (s)})} = \frac{1}{2}[/tex].
Rearrange this equation to obtain:
[tex]\displaystyle n(\mathrm{Cu(NO_3)_2\, (aq)}) = \frac{1}{2} \, n(\mathrm{Ag\, (s)})[/tex].
In other words, [tex]n(\rm Cu(NO_3)_2\; (aq))[/tex] (the number of moles of [tex]\rm Cu(NO_3)_2\; (aq)[/tex] that is produced) can be found from [tex]n(\mathrm{Ag\, (s)})[/tex] (the number of moles of [tex]\rm Ag\, (s)[/tex] that is produced.)
On the other hand, [tex]n(\rm Ag\; (aq))[/tex] can be found from [tex]m(\rm Ag\; (s))[/tex] (the mass of [tex]\rm Ag\, (s)[/tex] that is produced) using [tex]M(\rm Ag)[/tex] (the formula mass of [tex]\rm Ag[/tex].)
Look up relative atomic mass data of [tex]\mathrm{Ag}[/tex] on a modern periodic table:
- [tex]\rm Ag[/tex]: [tex]107.868[/tex].
Therefore, the relative atomic mass of [tex]\rm Ag\, (s)[/tex] would be:
[tex]M(\rm Ag) = 107.868\; \rm g \cdot mol^{-1}[/tex].
Calculate [tex]n(\rm Ag\; (s))[/tex]:
[tex]\begin{aligned}& n({\rm Ag\, (s)}) \\ &= \frac{m(\mathrm{Ag\, (s)})}{M(\mathrm{Ag})}\\ &=\frac{2.60\; \rm g}{107.868 \; \rm g \cdot mol^{-1}} \\ &\approx 0.0241035\; \rm mol \end{aligned}[/tex].
Calculate [tex]n(\rm Cu(NO_3)_2\; (aq))[/tex]:
[tex]\displaystyle n(\mathrm{Cu(NO_3)_2\, (aq)}) = \frac{1}{2} \, n(\mathrm{Ag\, (s)}) \approx 0.0121\; \rm mol[/tex].
