Chemistry • Year 11 • Module 2 • Lesson 13
Limiting Reagents & Theoretical Yield
Build HSC Band 5–6 technique on multi-part limiting-reagent problems, experimental design for yield investigations, and evaluating flawed student reasoning.
1. Full limiting-reagent problem — ammonia synthesis (Band 4–5)
10 marks Band 4–5
Scenario. The Haber process is an industrial reaction used in Australia to manufacture fertiliser: N2(g) + 3H2(g) → 2NH3(g). A chemist has 28.0 g of N2 and 9.00 g of H2. (N = 14.007, H = 1.008)
Q1. In your response:
- Identify the limiting reagent using the full n ÷ coefficient comparison method. Show all working.
- Calculate the theoretical yield of NH3 in grams.
- Calculate the mass of the excess reagent remaining after the reaction is complete.
- State which reactant is in excess and by how many grams.
2. Evaluate a student’s reasoning — combustion of methane (Band 5–6)
8 marks Band 5–6
Student’s claim.
“In the reaction CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), I have 16.0 g of CH4 and 64.0 g of O2. To find the theoretical yield of CO2, I should use whichever reagent gives me the larger amount of product — that must be the maximum. O2 gives more CO2, so I’ll use O2 and get m(CO2) = 88.0 g.” (C = 12.011, H = 1.008, O = 15.999)
Q2. Evaluate the student’s reasoning. In your response:
- Identify the limiting reagent using the correct method. Show your working.
- Calculate the correct theoretical yield of CO2.
- Calculate the yield the student would get using the excess reagent. Show this working.
- Explain, using the concept of “reaction-units” (n ÷ coefficient), why using the excess reagent overestimates the yield and is fundamentally wrong.
- Evaluate whether the student’s strategy (“use whichever gives the larger yield”) could ever give the correct answer by coincidence, and under what condition.
Q1 — Sample Band 5 response (10 marks), annotated
Step 1 — Moles:
n(N2) = 28.0 ÷ 28.014 = 0.9995 mol. n(H2) = 9.00 ÷ 2.016 = 4.464 mol. [1 mark for both correct]
Step 2 — n ÷ coefficient comparison:
N2: 0.9995 ÷ 1 = 0.9995. H2: 4.464 ÷ 3 = 1.488. N2 has the smaller quotient (0.9995 < 1.488). N2 is the limiting reagent. [1 mark for both quotients; 1 mark for correct LR identification with comparison]
Step 3 — Theoretical yield of NH3:
Ratio N2:NH3 = 1:2. n(NH3) = 0.9995 × 2 = 1.999 mol. MM(NH3) = 14.007 + 3(1.008) = 17.031 g/mol. m(NH3) = 1.999 × 17.031 = 34.0 g. [1 mark for correct n(NH3); 1 mark for correct mass]
Step 4 — Excess H2 remaining:
Ratio N2:H2 = 1:3. n(H2 consumed) = 0.9995 × 3 = 2.999 mol. n(H2 remaining) = 4.464 − 2.999 = 1.465 mol. m(H2 remaining) = 1.465 × 2.016 = 2.95 g of H2 remains unreacted. [1 mark for correct consumed; 1 mark for correct remaining mass]
Statement: H2 is in excess; 2.95 g of H2 remains after N2 is fully consumed. [1 mark]
Marking criteria (10 marks): 1 = both n values correct; 1 = both n ÷ coeff quotients correct; 1 = correct LR identified with explicit comparison; 1 = correct n(NH3); 1 = correct m(NH3) = 34.0 g; 1 = correct n(H2 consumed); 1 = correct n(H2 remaining); 1 = correct m(H2 remaining) = 2.95 g; 1 = correct identification of H2 as excess; 1 = clear, systematic layout throughout.
Q2 — Sample Band 6 response (8 marks), annotated
Identifying the limiting reagent (correct method):
n(CH4) = 16.0 ÷ 16.043 = 0.9973 mol. n(O2) = 64.0 ÷ 31.998 = 2.000 mol. CH4: 0.9973 ÷ 1 = 0.9973. O2: 2.000 ÷ 2 = 1.000. CH4 has the smaller quotient (0.9973 < 1.000). CH4 is the limiting reagent. [1 mark]
Correct theoretical yield of CO2:
Ratio CH4:CO2 = 1:1. n(CO2) = 0.9973 mol. MM(CO2) = 12.011 + 2(15.999) = 44.009 g/mol. m(CO2) = 0.9973 × 44.009 = 43.9 g. [1 mark for n; 1 mark for mass]
Student’s incorrect yield using O2:
Ratio O2:CO2 = 2:1. n(CO2) = 2.000 × (1÷2) = 1.000 mol. m(CO2) = 1.000 × 44.009 = 44.0 g (not 88.0 g as claimed — the student also made an arithmetic error). This is larger than the correct yield of 43.9 g. [1 mark for showing the calculation]
Why using the excess reagent overestimates:
The n ÷ coefficient quotient for O2 (1.000) is larger than for CH4 (0.9973), meaning O2 has more “reaction-units” available than CH4 can react with. When CH4 runs out, the reaction stops — some O2 remains unconsumed. Using O2’s moles assumes all the O2 reacts, which is impossible if CH4 is already exhausted. The theoretical yield must always be calculated from the reactant that limits the reaction. [1 mark for reaction-units reasoning; 1 mark for explaining that reaction stops at LR]
Evaluating whether the strategy could ever give the correct answer:
The student’s strategy “use whichever gives the larger yield” is fundamentally backwards. The correct yield always comes from the smaller quotient (the LR), which always gives the lower yield. The student’s approach would never coincidentally give the right answer, except in the special case where both reactants are mixed in exactly stoichiometric proportions (n ÷ coeff values are equal) — at that point both reagents are limiting simultaneously and both give the same yield. [1 mark for identifying stoichiometric-ratio special case; 1 mark for clear evaluative judgement that the strategy is always wrong unless stoichiometric]
Marking criteria (8 marks): 1 = correct LR identified with full comparison; 1 = correct n(CO2) from LR; 1 = correct m(CO2) = 43.9 g; 1 = shows calculation of (larger, incorrect) yield if O2 used; 1 = uses reaction-units concept to explain why excess reagent overestimates; 1 = explains reaction stops when LR is consumed; 1 = evaluates when strategy coincidentally works (stoichiometric ratio); 1 = reaches an explicit evaluative judgement (strategy is always wrong in non-stoichiometric cases).