Question

Copper reacts with silver nitrate through single replacement.

If 2.60 g of silver is produced from the reaction, how many moles of copper(II) nitrate are also produced?

Answer 1

Approximately
`0.121\; \rm mol`.

Step-by-step explanation:

Balance the equation

`\rm Cu\; (s)` reacts with
`\rm AgNO_3\; (aq)` to produce
`\rm Ag\; (s)` and
`\rm Cu(NO_3)_2\; (aq)`.

`\rm ? \; Cu\, (s) + ?\; AgNO_3\, (aq) \to ?\; Ag\, (s) + ?\; Cu(NO_3)_2\, (aq)`.

To balance this equation, start by setting the coefficient of the most complex species to
`1`. For example,
`\rm Cu(NO_3)_2\, (aq)` has more atoms in each of its formula unit than any other species in this reaction. Let the coefficient of

`\rm ? \; Cu\, (s) + ?\; AgNO_3\, (aq) \to ?\; Ag\, (s) + 1\; Cu(NO_3)_2\, (aq)`.

That formula unit of
`\rm Cu(NO_3)_2\, (aq)` would include:

• 
  `1 * 1 = 1`
  `\rm Cu` atom,
• 
  `1 * 2= 2`
  `\rm N` atoms, and
• 
  `1 * 2 * 3 = 6`
  `\rm O` atoms.

The other product,
`\rm Ag\; (s)`, contains neither
`\rm Cu` atoms nor
`\rm N` atoms. Therefore, the product side would include exactly

• one
  `\rm Cu` atom,
• two
  `\rm N` atoms, and
• six
  `\rm O` atoms.

Atoms are conserved in a chemical reaction. The reactant side shall also include:

• 
  `1 * 1 = 1`
  `\rm Cu` atom,
• 
  `1 * 2= 2`
  `\rm N` atoms, and
• 
  `1 * 2 * 3 = 6`
  `\rm O` atoms.

Among the reactants,
`\rm AgNO_3\, (aq)` is the only source of
`\rm N` atoms. Each formula unit of

Similarly, the coefficient of
`\rm Cu` would be
`1/ 1 = 1`.

`\rm 1 \; Cu\, (s) + 1\; AgNO_3\, (aq) \to ?\; Ag\, (s) + 1\; Cu(NO_3)_2\, (aq)`.

There would be exactly
`2`
`\rm Ag` atom on the reactant side (from
`\rm AgNO_3\, (aq)`.) The products should also include

Hence the equation:

`\rm 1 \; Cu\, (s) + 1\; AgNO_3\, (aq) \to 2\; Ag\, (s) + 1\; Cu(NO_3)_2\, (aq)`.

Calculate n(Cu(NO₃)₂ (aq))

The ratio between the coefficients of
`\rm Cu(NO_3)_2\; (aq)` and
`\rm Ag\, (s)` is:

`\displaystyle \frac{n(\mathrm{Cu(NO_3)_2\, (aq)})}{n(\mathrm{Ag\, (s)})} = (1)/(2)`.

Rearrange this equation to obtain:

`\displaystyle n(\mathrm{Cu(NO_3)_2\, (aq)}) = (1)/(2) \, n(\mathrm{Ag\, (s)})`.

In other words,
`n(\rm Cu(NO_3)_2\; (aq))` (the number of moles of
`\rm Cu(NO_3)_2\; (aq)` that is produced) can be found from
`n(\mathrm{Ag\, (s)})` (the number of moles of
`\rm Ag\, (s)` that is produced.)

On the other hand,
`n(\rm Ag\; (aq))` can be found from
`m(\rm Ag\; (s))` (the mass of
`\rm Ag\, (s)` that is produced) using
`M(\rm Ag)` (the formula mass of
`\rm Ag`.)

Look up relative atomic mass data of
`\mathrm{Ag}` on a modern periodic table:

• 
  `\rm Ag`:
  `107.868`.

Therefore, the relative atomic mass of
`\rm Ag\, (s)` would be:

`M(\rm Ag) = 107.868\; \rm g \cdot mol^(-1)`.

Calculate
`n(\rm Ag\; (s))`:

`\begin{aligned}& n({\rm Ag\, (s)}) \\ &= \frac{m(\mathrm{Ag\, (s)})}{M(\mathrm{Ag})}\\ &=(2.60\; \rm g)/(107.868 \; \rm g \cdot mol^(-1)) \\ &\approx 0.0241035\; \rm mol \end{aligned}`.

Calculate
`n(\rm Cu(NO_3)_2\; (aq))`:

`\displaystyle n(\mathrm{Cu(NO_3)_2\, (aq)}) = (1)/(2) \, n(\mathrm{Ag\, (s)}) \approx 0.0121\; \rm mol`.