Chemistry • Year 11 • Module 2 • Lesson 13

Limiting Reagents & Theoretical Yield

Apply the five-step LR method to calculate theoretical yield, identify excess reagents, and interpret real reaction data across a range of stoichiometric contexts.

Apply · Data & Reasoning

1. Identify the limiting reagent from the data table

Three reactions are listed below. For each reaction, the moles of each reactant have been calculated for you. Complete the table by dividing each by its coefficient and identifying the limiting reagent. 9 marks (3 each)

Reaction	Reactant A n & coefficient	Reactant B n & coefficient
Fe + S → FeS	n(Fe) = 0.200 mol; coeff = 1	n(S) = 0.350 mol; coeff = 1
2Na + Cl₂ → 2NaCl	n(Na) = 0.500 mol; coeff = 2	n(Cl₂) = 0.310 mol; coeff = 1
N₂ + 3H₂ → 2NH₃	n(N₂) = 0.400 mol; coeff = 1	n(H₂) = 0.900 mol; coeff = 3

Stuck? Revisit the bar diagram in the lesson showing how to compare H₂ and O₂ as reaction-units. The smaller n ÷ coeff is the LR.

2. Interpret a bar graph — reaction-unit comparison

The bar graph below shows the n ÷ coefficient values for two reactants in the reaction: 2Al + 3Cl₂ → 2AlCl₃. 6 marks

Figure 2.1. n ÷ coefficient comparison for 2Al + 3Cl₂ → 2AlCl₃ with n(Al) = 0.400 mol and n(Cl₂) = 0.300 mol.

2.1 Using the graph, identify the limiting reagent and explain your reasoning. (2 marks)

2.2 Using the original data (n(Al) = 0.400 mol, n(Cl₂) = 0.300 mol), show the full working to calculate the theoretical yield of AlCl₃ in grams. (Al = 26.982, Cl = 35.453) (3 marks)

2.3 Calculate the mass of Al remaining after the reaction. (1 mark)

Stuck? Revisit Worked Example 4 (2Al + 3Cl₂) in the lesson for a fully worked version of this reaction type.

3. Predict and justify — the sandwich-making analogy

A school canteen makes ham-and-cheese rolls. Each roll requires 2 slices of bread and 1 serving of filling. The canteen has 18 slices of bread and 11 servings of filling. 5 marks

3.1 Using the same method as a limiting-reagent calculation (divide quantity by “coefficient”), identify which ingredient limits the number of rolls made. Show the comparison. (3 marks)

3.2 How many rolls can be made? How much of the excess ingredient is left over? (2 marks)

Stuck? The lesson’s sandwich analogy (10 bread, 3 filling → filling is limiting) uses the same logic as n ÷ coefficient.

4. Identify and correct the student’s error

A student is asked: “8.00 g of Na reacts with 12.0 g of Cl₂. Which is the limiting reagent? (2Na + Cl₂ → 2NaCl; Na = 22.990, Cl = 35.453)” Their working is shown below. 6 marks

Student’s working:

n(Na) = 8.00 ÷ 22.990 = 0.3480 mol
n(Cl₂) = 12.0 ÷ 70.906 = 0.1692 mol
Cl₂ has fewer moles (0.1692 < 0.3480)
∴ Cl₂ is the limiting reagent.

4.1 Identify the error in the student’s method. Explain what they should have done instead. (2 marks)

4.2 Show the correct working and state which reactant is actually the limiting reagent. (2 marks)

4.3 Using the correct limiting reagent, calculate the theoretical yield of NaCl in grams. (MM(NaCl) = 58.443 g/mol) (2 marks)

Stuck? Revisit the Common Mistakes box and Worked Example 2 (Na + Cl₂) in the lesson.

Answers — Do not peek before attempting

Q1 — Data table

Fe + S → FeS: Fe: 0.200 ÷ 1 = 0.200; S: 0.350 ÷ 1 = 0.350. LR = Fe (0.200 < 0.350).

2Na + Cl₂ → 2NaCl: Na: 0.500 ÷ 2 = 0.250; Cl₂: 0.310 ÷ 1 = 0.310. LR = Na (0.250 < 0.310).

N₂ + 3H₂ → 2NH₃: N₂: 0.400 ÷ 1 = 0.400; H₂: 0.900 ÷ 3 = 0.300. LR = H₂ (0.300 < 0.400).

Marking criteria (3 marks each): 1 mark for both correct n ÷ coeff quotients; 1 mark for correct identification of LR; 1 mark for a clear comparison statement.

Q2.1 — LR from graph

Cl₂ is the limiting reagent because its n ÷ coefficient value (0.10) is smaller than Al’s value (0.20). The bar for Cl₂ is shorter, meaning Cl₂ has fewer “reaction-units” relative to its role in the equation.

Q2.2 — Theoretical yield of AlCl₃

LR = Cl₂; n(Cl₂) = 0.300 mol. Ratio Cl₂:AlCl₃ = 3:2. n(AlCl₃) = 0.300 × (2÷3) = 0.200 mol. MM(AlCl₃) = 26.982 + 3(35.453) = 133.34 g/mol. m(AlCl₃) = 0.200 × 133.34 = 26.7 g.

Q2.3 — Al remaining

n(Al consumed) = n(Cl₂) × (2÷3) = 0.300 × 0.667 = 0.200 mol. n(Al remaining) = 0.400 − 0.200 = 0.200 mol. m(Al remaining) = 0.200 × 26.982 = 5.40 g.

Q3.1 — Sandwich analogy

Bread: 18 ÷ 2 = 9 (reaction-units). Filling: 11 ÷ 1 = 11 (reaction-units). Bread has the smaller quotient (9 < 11), so bread is the limiting ingredient.

Marking criteria: 1 mark for correct “quotients” (18/2 and 11/1); 1 mark for correct comparison; 1 mark for identifying bread as limiting with reason.

Q3.2 — Rolls made and excess

9 rolls can be made (limited by bread). Filling used = 9 × 1 = 9 servings. Filling remaining = 11 − 9 = 2 servings of filling left over.

Q4.1 — Error identification

The student compared raw moles directly without dividing by stoichiometric coefficients. In 2Na + Cl₂ → 2NaCl, Na has a coefficient of 2 while Cl₂ has a coefficient of 1. The correct method is to compute n ÷ coefficient for each reactant before comparing.

Q4.2 — Correct working

Na: 0.3480 ÷ 2 = 0.1740. Cl₂: 0.1692 ÷ 1 = 0.1692. Cl₂ has the smaller quotient (0.1692 < 0.1740). Cl₂ is the limiting reagent. (The student’s conclusion was actually correct; their method was wrong but happened to give the right answer in this case. In other ratios this shortcut fails.)

Q4.3 — Theoretical yield of NaCl

LR = Cl₂; n(Cl₂) = 0.1692 mol. Ratio Cl₂:NaCl = 1:2. n(NaCl) = 0.1692 × 2 = 0.3384 mol. m(NaCl) = 0.3384 × 58.443 = 19.8 g.

Marking criteria (2 marks): 1 mark for correct moles of NaCl from LR; 1 mark for correct mass calculation.