Core idea
The phrase “which terms characterize the product of the reaction” is most commonly answered in general chemistry by the quantitative terms that describe how much product is formed compared with the maximum predicted by stoichiometry. The central product-characterizing terms are theoretical yield, actual yield, and percent yield.
Key terms
- Theoretical yield: the maximum amount of product possible, calculated from the balanced chemical equation assuming complete conversion of the limiting reactant.
- Actual yield: the amount of product obtained experimentally (measured mass or moles).
- Percent yield: a comparison of actual to theoretical yield, expressed as a percentage: \( \text{Percent yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \).
How these terms are determined from reaction stoichiometry
- Balance the equation to fix mole ratios between reactants and products.
- Convert given reactant amounts (often grams) to moles using molar masses.
- Identify the limiting reactant by checking which reactant produces the smallest moles of product.
- Compute the theoretical yield of the desired product from the limiting reactant using the mole ratio, then convert to grams if needed.
- Use the measured mass as the actual yield and compute percent yield with the formula above.
Worked example
Reaction: \(2\,\text{Al} + 3\,\text{Cl}_2 \rightarrow 2\,\text{AlCl}_3\)
Suppose the reaction starts with 10.0 g Al and 20.0 g Cl2. Molar masses (g/mol): Al \(= 26.98\), Cl2 \(= 70.90\), AlCl3 \(= 133.33\). After the reaction, the collected AlCl3 is 22.5 g (actual yield).
1) Convert reactants to moles
\[ n(\text{Al}) = \frac{10.0}{26.98} = 0.3706\ \text{mol} \] \[ n(\text{Cl}_2) = \frac{20.0}{70.90} = 0.2822\ \text{mol} \]2) Find the limiting reactant
The balanced equation requires \(3\) mol Cl2 per \(2\) mol Al, so the required ratio is \( \dfrac{n(\text{Cl}_2)}{n(\text{Al})} = \dfrac{3}{2} = 1.5\).
\[ n(\text{Cl}_2)\ \text{needed for}\ 0.3706\ \text{mol Al} = 0.3706 \times \frac{3}{2} = 0.5559\ \text{mol} \]Since only \(0.2822\) mol Cl2 is available, Cl2 is the limiting reactant.
3) Theoretical yield of AlCl3
From \(3\,\text{Cl}_2 \rightarrow 2\,\text{AlCl}_3\), the mole ratio is \( \dfrac{2}{3}\).
\[ n(\text{AlCl}_3)_{\text{theo}} = 0.2822 \times \frac{2}{3} = 0.1881\ \text{mol} \] \[ m(\text{AlCl}_3)_{\text{theo}} = 0.1881 \times 133.33 = 25.08\ \text{g} \]4) Percent yield
\[ \text{Percent yield} = \frac{22.5}{25.08}\times 100\% = 89.7\% \]| Quantity (product characterization) | Meaning | Value for the example (AlCl3) |
|---|---|---|
| Theoretical yield | Maximum predicted product from the limiting reactant | 25.08 g |
| Actual yield | Measured product obtained in the laboratory | 22.5 g |
| Percent yield | Efficiency of product formation relative to theoretical maximum | 89.7% |
Visualization
Common interpretation notes
- Percent yield above 100% often indicates impurities, incomplete drying, or measurement error rather than “extra” product.
- Theoretical yield depends on the limiting reactant; changing initial amounts can change which reactant limits and therefore changes the predicted product amount.
- If multiple products form, product characterization may also include selectivity or yield of a specific product, but the standard quantitative terms remain theoretical yield, actual yield, and percent yield.