Answer:
The theoretical yield of copper is approximately 1.65 grams.
Explanation:
Relative atomic mass data from a modern periodic table:
As a metal, aluminum is more reactive than copper. Aluminum will reduce the [tex]\text{Cu}^{2+}[/tex] ion in [tex]\text{CuCl}_2\;(aq)[/tex] to [tex]\text{Cu}\;(s)[/tex]. This process will form a positive aluminum ion, which will then combine with chloride ions in the solution. What will be the charge on each aluminum ion? Aluminum is in IUPAC Group 13 of the periodic table. Each aluminum atom contains three valence electrons. As a main group metal in the p-block, each atom will lose all three of its valence electrons to form [tex]\text{Al}^{3+}[/tex] ions with three positive charges. Each ion will combine with three [tex]\text{Cl}^{-}[/tex] ions to produce a species with the empirical formula [tex]\text{Al}\text{Cl}_3[/tex].
Reactants:
Products:
Let the coefficient in front of [tex]\text{AlCl}_3\;(aq)[/tex] be [tex]1[/tex].
[tex]\begin{array}{cccccccl}\text{Al}\;(s) & +& \text{CuCl}_2\;(aq) & \to & \text{AlCl}_3\;(aq) & + & \text{Cu}\;(s)\\ & & & & {\bf 1} & & &\begin{aligned}&\text{Assign "1" to the most}\\[-0.5em]&\text{complex compound.}\end{aligned} \\ {\bf 1}& &{\bf 3/2} & & 1 & & &\begin{aligned}&\text{Al and Cl atoms}\\[-0.5em]&\text{conserve.}\end{aligned}\end{array}[/tex][tex]\begin{array}{cccccccl}\phantom{\;\text{Al}\;(s)} & \phantom{+}& \phantom{\text{CuCl}_2\;(aq)} & \phantom{\to} & \phantom{\text{AlCl}_3\;(aq)} & \phantom{+} & \phantom{\text{Cu}\;(s)}\\[-1em]1& &{3/2} & & 1 & & {\bf 3/2} &\begin{aligned}&\text{Cu atoms shall also}\\[-0.5em]&\text{conserve.}\end{aligned} \\2& &{3} & & 2 & & {3} &\begin{aligned}&\text{Multiply all}\\[-0.5em]&\text{coefficients by two}\\[-0.5em]&\text{to eliminate fractions.}\end{aligned}\end{array}[/tex].
Hence the balanced equation:
[tex]2 \;\text{Al} \;(s) + 3 \;\text{CuCl}_2\;(aq) \to 2\;\text{AlCl}_3\;(aq) + 3\;\text{Cu}\;(s)[/tex].
Assume that [tex]\text{CuCl}_2[/tex] is the limiting reactant.
Formula mass of [tex]\text{CuCl}_2[/tex]:
[tex]M(\text{CuCl}_2) = 63.546 + 2 \times 35.45 = 134.446\;\text{g}\cdot\text{mol}^{-1}[/tex].
Number of moles of [tex]\text{CuCl}_2[/tex] available:
[tex]\displaystyle n = \frac{m}{M} = \frac{3.5}{134.446} = 0.0260328\;\text{mol}[/tex].
The ratio between the coefficient in front of [tex]\text{CuCl}_2[/tex] is the same as the coefficient in front of [tex]\text{Cu}[/tex].
[tex]\displaystyle \frac{n(\text{Cu})}{n(\text{CuCl}_2)} = \frac{\text{Coefficient in front of } n(\text{Cu})}{\text{Coefficient in front of } n(\text{CuCl}_2)} = \frac{3}{3} = 1[/tex].
[tex]n(\text{Cu}) = 1 \cdot n(\text{CuCl}_2) = 0.0260328\;\text{mol}[/tex].
Mass of copper that is expected to be produced if [tex]\text{CuCl}_2[/tex] is the limiting reactant:
[tex]m(\text{Cu}) = n(\text{Cu}) \cdot M(\text{Cu}) = 0.0260328 \times 63.546 =1.65\;\text{g}[/tex].
Assume that [tex]\text{Al}[/tex] is the limiting reactant.
The methods are similar. Try the steps above yourself.
Formula mass of [tex]\text{Al}[/tex]:
[tex]M(\text{CuCl}_2) = 26.982\;\text{g}\cdot\text{mol}^{-1}[/tex].
Number of moles of [tex]\text{Al}[/tex] available:
[tex]\displaystyle n = \frac{m}{M} = \frac{0.5}{26.982} = 0.0185\;\text{mol}[/tex].
The ratio between the coefficient in front of [tex]\text{CuCl}_2[/tex] is [tex]3/2[/tex] times the coefficient in front of [tex]\text{Al}[/tex].
[tex]\displaystyle \frac{n(\text{Cu})}{n(\text{Al})} = \frac{\text{Coefficient in front of } n(\text{Cu})}{\text{Coefficient in front of } n(\text{Al})} = \frac{3}{2}[/tex].
[tex]\displaystyle n(\text{Cu}) = \frac{3}{2} \cdot n(\text{Al}) = 0.0277963\;\text{mol}[/tex].
Mass of copper that is expected to be produced if [tex]\text{Al}[/tex] is the limiting reactant:
[tex]m(\text{Cu}) = n(\text{Cu}) \cdot M(\text{Cu}) = 0.0277963 \times 63.546 =1.77\;\text{g}[/tex].
[tex]1.65\;\text{g} < 1.77\;\text{g}[/tex]. The first assumption is valid. [tex]\text{CuCl}_2[/tex] will run out before all [tex]0.5\;\text{g}[/tex] of aluminum are consumed, Only [tex]1.65\;\text{g}[/tex] of copper will be produced.
