Stoichiometry notes: limiting reactant and theoretical yield (lecture-style)

Conceptual Foundations

In chemical reactions, the coefficients in a balanced equation correspond to moles, not directly to countable items. The kitchen analogy helps: B u = butter, S u = sugar, SQ = shortbread squares. Coefficients like 1, 1/2, 2, 25 represent measurable quantities in cooking, but in chemistry they map to mole quantities.
The key idea: every run of a reaction has a limited amount of one reactant. This is the limiting reactant. Once that reactant is consumed, the reaction stops, even if other reactants remain in excess.
Why identify the limiting reactant first? The theoretical yield (the maximum amount of product that could be formed) assumes complete conversion of the limiting reactant. If you don’t know which reactant is limiting, you can’t correctly compute the theoretical yield.
Common pitfall: you cannot determine the limiting reactant simply by comparing the given masses or grams. You must convert masses to moles using molar masses and compare to the stoichiometric coefficients.
The concept of theoretical yield vs. actual yield:
- Theoretical yield is the amount of product formed if the reaction goes to completion with the limiting reactant fully consumed.
- In real experiments, the actual yield can be lower due to side reactions, incomplete reactions, or losses.

How to reason with a balanced equation

General form: $a\,A + b\,B \rightarrow c\,C$ where $a$, $b$, $c$ are the stoichiometric coefficients.
Step-by-step approach:
- Step 1: Convert the given masses to moles:
 $nA = \frac{mA}{MA}, \quad nB = \frac{mB}{MB}$
- Step 2: Determine which reactant is limiting by comparing the available mole ratios to the required ratio $a:b$:
 $\text{extent } \xi = \min\left( \frac{nA}{a}, \frac{nB}{b} \right)$
- Step 3: Convert the extent to moles of product formed:
 $n_C^{\text{theo}} = c \cdot \xi$
- Step 4: Convert to mass of product:
 $mC^{\text{theo}} = nC^{\text{theo}} \cdot M_C$
Important note: When one reactant is limiting, the other reactants may remain in excess after the reaction.

Worked Examples

Example 1: Ammonia and Oxygen to Nitric Oxide (NO)

Reaction (balanced):
$4\,\mathrm{NH3} + 5\,\mathrm{O2} \rightarrow 4\,\mathrm{NO} + 6\,\mathrm{H_2O}$
Given: 56.8 g of NH3; O2 in excess is assumed (NO is the product of interest).
Molar masses (approximate):
- $M{\mathrm{NH3}} \approx 17.03\ \mathrm{g\,mol^{-1}}$
- $M_{\mathrm{NO}} \approx 30.01\ \mathrm{g\,mol^{-1}}$
Step 1: Convert NH3 to moles:
$n{\mathrm{NH3}} = \frac{56.8}{17.03} \approx 3.34\ \text{mol}.$
Step 2: Stoichiometry for NO production (4 NH3 -> 4 NO gives a 1:1 mole ratio for NO and NH3 in this balanced equation):
$n{\mathrm{NO}}^{\text{theo}} = n{\mathrm{NH_3}} \;\text{(since }4:4\text{)} \approx 3.34\ \text{mol}.$
Step 3: Convert NO to mass:
$m{\mathrm{NO}}^{\text{theo}} = n{\mathrm{NO}}^{\text{theo}} \cdot M_{\mathrm{NO}} \approx 3.34 \times 30.01 \approx 100\ \text{g}.$
Conclusion: With O2 in excess, the theoretical yield of NO is about 100 g; NH3 is the limiting reactant in this setup.

Note: This example also illustrates the idea that the theoretical yield can be computed by converting the limiting reactant to the product via stoichiometry, then converting moles to grams.

Example 2: Iron and Chlorine to Ferric Chloride (FeCl3)

Reaction (balanced):
$2\,\mathrm{Fe} + 3\,\mathrm{Cl2} \rightarrow 2\,\mathrm{FeCl3}$
Given: 10.0 g Fe; assume FeCl3 is the product of interest.
Molar masses (approximate):
- $M_{\mathrm{Fe}} \approx 55.85\ \mathrm{g\,mol^{-1}}$
- $M{\mathrm{FeCl3}} \approx 162.20\ \mathrm{g\,mol^{-1}}$
Step 1: Convert Fe to moles:
$n_{\mathrm{Fe}} = \frac{10.0}{55.85} \approx 0.179\ \text{mol}.$
Step 2: Stoichiometry: 2 Fe produce 2 FeCl3, so mole ratio Fe:FeCl3 is 1:1. Therefore:
$n{\mathrm{FeCl3}}^{\text{theo}} = n_{\mathrm{Fe}} \approx 0.179\ \text{mol}.$
Step 3: Convert FeCl3 to mass:
$m{\mathrm{FeCl3}}^{\text{theo}} = 0.179 \times 162.20 \approx 29.0\ \text{g}.$
Conclusion: If Fe is the limiting reagent, the theoretical yield of FeCl3 is about 29.0 g.

Example 3: Bromine (Br2) Scenario (conceptual)

Given: 30 g Br2; atomic/molar considerations:
- Atomic weight Br ≈ 79.9 g/mol, so Br2 molar mass is
 $M{\mathrm{Br2}} = 2 \times 79.9 \approx 159.8\ \mathrm{g\,mol^{-1}}.$
Step 1: Convert Br2 to moles:
$n{\mathrm{Br2}} = \frac{30}{159.8} \approx 0.188\ \text{mol}.$
Step 2: In a generic reaction of Br2 with some reactant X (balanced as $a\,\mathrm{Br_2} + b\,\mathrm{X} \rightarrow \cdots$), determine the limiting reactant by comparing the required moles to the available moles: the reagent for which the required amount exceeds what is present is the limiting reactant.
Step 3: To obtain the theoretical yield of the product, use the stoichiometry to convert the limiting reagent moles to product moles, then to grams using the product molar mass, exactly as in the previous examples.
Note: The transcript emphasizes that you compare the mole amounts relative to the stoichiometric coefficients, not simply compare gram amounts, to identify the limiting reactant.

Quick-reference: key formulas

For a reaction $a\,A + b\,B \rightarrow c\,C$:
- Moles from mass: $nA = \frac{mA}{MA}, \quad nB = \frac{mB}{MB}$
- Limiting extent: $\xi = \min\left( \frac{nA}{a}, \frac{nB}{b} \right)$
- Theoretical product moles: $n_C^{\text{theo}} = c \cdot \xi$
- Theoretical product mass: $mC^{\text{theo}} = nC^{\text{theo}} \cdot M_C$
Useful conversions:
- Mass to moles: $n = \dfrac{m}{M}$
- Molar mass of Br2: $M{\mathrm{Br2}} = 2 \times 79.9 \approx 159.8\ \mathrm{g\,mol^{-1}}$

Practical takeaways

Always identify the limiting reactant before calculating the theoretical yield.
Use molar masses to convert masses to moles, then apply stoichiometry to map to product moles, and finally convert to product mass.
Do not judge limiting reagents by mass alone; convert to moles and use the balanced equation to compare required mole ratios.
Real-world relevance: this approach underpins chemical manufacturing, materials synthesis, and environmental calculations where yield predictions matter for cost and safety.