Problem
The reaction is \(\mathrm{2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)}\). A mixture of \(10.0\,\mathrm{g}\) Al and \(20.0\,\mathrm{g}\) \(\mathrm{Cl_2}\) produces an actual yield of \(18.5\,\mathrm{g}\) \(\mathrm{AlCl_3}\). Determine the percent yield.
Definition: \(\text{percent yield}=\dfrac{\text{actual yield}}{\text{theoretical yield}}\times 100\%\).
Plan: identify the limiting reactant, compute theoretical yield of \(\mathrm{AlCl_3}\), then apply the percent yield formula.
Solution
1) Convert given masses to moles
Use molar masses (in \(\mathrm{g\,mol^{-1}}\)): \(M(\mathrm{Al})=26.98\), \(M(\mathrm{Cl_2})=70.90\), \(M(\mathrm{AlCl_3})=26.98+3(35.45)=133.33\).
2) Determine the limiting reactant
The balanced equation requires \(3\) mol \(\mathrm{Cl_2}\) for every \(2\) mol \(\mathrm{Al}\). Compare the available \(\mathrm{Cl_2}\) to what would be required to consume all Al:
Only \(0.2822\,\mathrm{mol}\) \(\mathrm{Cl_2}\) is available, which is less than \(0.5559\,\mathrm{mol}\). Therefore, \(\mathrm{Cl_2}\) is the limiting reactant.
3) Compute the theoretical yield of \(\mathrm{AlCl_3}\)
From the stoichiometry \(3\,\mathrm{mol}\,\mathrm{Cl_2}\rightarrow 2\,\mathrm{mol}\,\mathrm{AlCl_3}\),
Convert theoretical moles of product to mass:
4) Calculate percent yield
The actual yield is \(18.5\,\mathrm{g}\). Apply the percent yield formula:
| Quantity | Value | How it was obtained |
|---|---|---|
| \(n(\mathrm{Al})\) | \(0.3706\,\mathrm{mol}\) | \(10.0/26.98\) |
| \(n(\mathrm{Cl_2})\) | \(0.2822\,\mathrm{mol}\) | \(20.0/70.90\) |
| Limiting reactant | \(\mathrm{Cl_2}\) | \(0.2822<0.5559\) (needed to consume all Al) |
| Theoretical yield | \(25.08\,\mathrm{g}\ \mathrm{AlCl_3}\) | \(\frac{2}{3}n(\mathrm{Cl_2})\cdot 133.33\) |
| Actual yield | \(18.5\,\mathrm{g}\ \mathrm{AlCl_3}\) | Measured/isolated mass |
| Percent yield | \(\approx 73.8\%\) | \(\frac{18.5}{25.08}\times 100\%\) |
Visualization
The diagram compares the theoretical yield (maximum possible from the limiting reactant) to the actual yield (mass isolated), and the ratio is the percent yield.
Final answer
The limiting reactant is \(\mathrm{Cl_2}\), the theoretical yield is \(25.08\,\mathrm{g}\) \(\mathrm{AlCl_3}\), and the percent yield is \(\frac{18.5}{25.08}\times 100\%\approx 73.8\%\).