Most real reactions don't have reactants in perfect stoichiometric proportions — one runs out first and stops the reaction. Identifying which reactant is the limiting reagent, and calculating the maximum possible yield, are among the most tested skills in HSC Chemistry.
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When you make sandwiches, having more bread than fillings means you'll run out of filling first — the filling limits how many sandwiches you can make. In chemistry, how would you decide which of two reactants "runs out first" and stops the reaction? What information would you need beyond just the masses given?
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📚 Core Content
Wrong: Gases have no mass because they float.
Right: Gases have mass; their density is just much lower than solids and liquids.
When two reactants are mixed in non-stoichiometric amounts, one will be completely consumed before the other. The reactant that runs out first is the limiting reagent (LR) — it limits the amount of product that can form. The other reactant is the excess reagent — some of it remains unreacted when the reaction stops.
The key step is dividing each reactant's moles by its coefficient in the balanced equation. This normalises the amounts — it tells you how many "reaction-units" of each reactant you have. The reactant with the smaller normalised value is the limiting reagent.
The theoretical yield is the maximum mass of product that can form, calculated from the moles of the limiting reagent. Once you have identified the LR, the yield calculation is just the standard 4-step stoichiometry method applied to the LR.
You can also calculate how much of the excess reagent is left over after the reaction. Find how much of the excess reagent is consumed (using the LR moles and the ratio), then subtract from the original amount.
🧮 Worked Examples
🧪 Activities
1 4.00 g of Fe reacts with 4.00 g of S. Fe + S → FeS. Identify the LR and calculate theoretical yield of FeS. (Fe = 55.845, S = 32.06)
2 6.00 g of H₂ reacts with 32.0 g of O₂. 2H₂ + O₂ → 2H₂O. Identify the LR and calculate m(H₂O). (H = 1.008, O = 15.999)
3 From Q2 above — calculate the mass of H₂ remaining after the reaction.
4 5.60 g of N₂ reacts with 1.50 g of H₂. N₂ + 3H₂ → 2NH₃. Identify the LR. (N = 14.007, H = 1.008)
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| Reaction | Amounts given | Find |
|---|---|---|
| 2Mg + O₂ → 2MgO | m(Mg) = 4.86 g m(O₂) = 3.20 g |
LR, m(MgO), m(excess remaining) |
| 4NH₃ + 5O₂ → 4NO + 6H₂O | m(NH₃) = 8.50 g m(O₂) = 16.0 g |
LR, m(NO) |
Mg=24.305, O=15.999, N=14.007, H=1.008
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At the start of this lesson, you thought about how you would decide which of two reactants "runs out first" in a chemical reaction and what information you would need beyond just the masses.
The key insight is that you cannot compare masses directly — you must convert to moles and then divide by each reactant's coefficient. The reactant with the smallest "moles ÷ coefficient" value is the limiting reagent, because it is the first to be fully consumed. The theoretical yield of any product must be calculated from the limiting reagent alone.
Reflect: how did your initial thinking compare to what you've learned?
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✍️ Short Answer
6. 5.40 g of Al reacts with 10.65 g of Cl₂. 2Al + 3Cl₂ → 2AlCl₃. (a) Show which reactant is the limiting reagent using the correct comparison method. (b) Calculate the theoretical yield of AlCl₃. (Al = 26.982, Cl = 35.453) 5 MARKS
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7. 16.0 g of CH₄ reacts with 64.0 g of O₂. CH₄ + 2O₂ → CO₂ + 2H₂O. (a) Identify the limiting reagent. (b) Calculate the theoretical yield of CO₂. (c) Calculate the mass of the excess reagent remaining after the reaction. (C=12.011, H=1.008, O=15.999) 6 MARKS
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8. A student says: "I mixed 10 g of each reactant, so neither one can be limiting — they're in equal amounts." Explain why this reasoning is incorrect and describe the correct method for identifying the limiting reagent. 3 MARKS
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9. 14.0 g of N₂ is mixed with 6.00 g of H₂ for the Haber process: N₂ + 3H₂ → 2NH₃. (a) Identify the limiting reagent using the correct method. (b) Calculate the theoretical yield of NH₃. (c) Calculate the mass of the excess reagent remaining after the reaction. (N = 14.007, H = 1.008) 6 MARKS
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10. In the reaction 4NH₃ + 5O₂ → 4NO + 6H₂O, a student is given 17.0 g of NH₃ and 48.0 g of O₂. She correctly identifies the limiting reagent and calculates the theoretical yield of NO. Another student argues: "We should use whichever reagent gives the larger amount of product, because that's the maximum." Evaluate this second student's reasoning and explain what is wrong with it. (N = 14.007, H = 1.008, O = 15.999) 4 MARKS
Calculate both possible yields, then explain the error in reasoning:
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Reaction 1 (2Mg + O₂ → 2MgO):
n(Mg) = 4.86÷24.305 = 0.2000; ÷2 = 0.1000 n(O₂) = 3.20÷31.998 = 0.1000; ÷1 = 0.1000Equal quotients — both consumed exactly (stoichiometric ratio). No excess remains.
n(MgO) = 0.2000 × 1 = 0.2000 mol; MM(MgO)=40.304; m(MgO) = 0.2000×40.304 = 8.06 gReaction 2 (4NH₃ + 5O₂ → 4NO + 6H₂O):
n(NH₃) = 8.50÷17.031 = 0.4992; ÷4 = 0.1248 n(O₂) = 16.0÷31.998 = 0.5001; ÷5 = 0.1000 O₂ has smaller quotient (0.1000 < 0.1248) → O₂ is LR Ratio O₂:NO = 5:4; n(NO) = 0.5001×(4÷5) = 0.4001 mol MM(NO) = 30.006; m(NO) = 0.4001×30.006 = 12.0 g1. B — H₂: 1.20÷3 = 0.40; N₂: 0.50÷1 = 0.50. H₂ has the smaller quotient → LR. Options A and C use the wrong comparison method.
2. C — Cl₂: 0.30÷3 = 0.10; Al: 0.30÷2 = 0.15. Cl₂ has the smaller quotient → LR. Option D has the division backwards (wrong which is smaller).
3. A — Theoretical yield uses LR moles × mole ratio to product. Using excess reagent moles gives a falsely high answer.
4. D — 35.7 g. n(H₂) = 4.00÷2.016 = 1.984 mol (LR); ratio H₂:H₂O = 1:1; n(H₂O) = 1.984; m = 1.984×18.015 = 35.7 g.
5. B — 0.26 g. n(H₂) = 4.00÷2.016 = 1.984 mol (LR). n(O₂ consumed) = 1.984×(1÷2) = 0.992 mol. n(O₂ initial) = 32.0÷31.998 = 1.000 mol. n(O₂ remaining) = 1.000−0.992 = 0.008 mol. m(O₂ remaining) = 0.008×31.998 = 0.26 g.
6. D — The student's conclusion (Fe is limiting) happens to be correct, but her method is wrong. Comparing raw moles without dividing by coefficients is invalid. The correct comparison is n÷coefficient: n(Fe)=5.58÷55.845=0.09992÷2=0.04996; n(Cl₂)=7.10÷70.906=0.1001÷3=0.03337. Cl₂ has the smaller quotient, making Cl₂ the actual limiting reagent — the opposite of what the student concluded.
7. A — This question tests whether students can work backwards and recognise when a problem has no solution. n(Cl₂) = 5.00÷70.906 = 0.07052 mol; ratio Cl₂:AlCl₃ = 3:2; n(AlCl₃) = 0.07052×(2÷3) = 0.04701 mol; m(AlCl₃) = 0.04701×133.34 = 6.27 g. Only 6.27 g of AlCl₃ can form from 5.00 g of Cl₂, regardless of how much Al is added. Option A correctly identifies this impossibility.
Q6 (5 marks):
(a) n(Al) = 5.40÷26.982 = 0.2001 mol; ÷2 = 0.1001 n(Cl₂) = 10.65÷70.906 = 0.1502 mol; ÷3 = 0.05006 Cl₂ has smaller quotient (0.0501 < 0.1001) → Cl₂ is the limiting reagent (b) Ratio Cl₂:AlCl₃ = 3:2; n(AlCl₃) = 0.1502×(2÷3) = 0.1001 mol MM(AlCl₃) = 133.34; m(AlCl₃) = 0.1001×133.34 = 13.4 gQ7 (6 marks):
(a) n(CH₄) = 16.0÷16.043 = 0.9973 mol; ÷1 = 0.9973 n(O₂) = 64.0÷31.998 = 2.000 mol; ÷2 = 1.000 CH₄ has smaller quotient (0.9973 < 1.000) → CH₄ is the LR (b) Ratio CH₄:CO₂ = 1:1; n(CO₂) = 0.9973 mol; MM(CO₂)=44.009; m = 0.9973×44.009 = 43.9 g (c) n(O₂ consumed) = 0.9973×(2÷1) = 1.995 mol; n(O₂ remaining) = 2.000−1.995 = 0.005 mol m(O₂ remaining) = 0.005×31.998 = 0.16 gQ8 (3 marks): Equal masses do not mean equal moles, because different substances have different molar masses. Even if the masses are identical, the number of moles of each reactant (n = m ÷ MM) will differ unless their molar masses happen to be equal. Furthermore, even equal moles does not guarantee stoichiometric equivalence — the coefficients in the balanced equation determine the ratio in which reactants are consumed. The correct method is to calculate moles of each reactant, divide each by its coefficient in the balanced equation, and identify the reactant with the smaller quotient as the limiting reagent.
Q9 (6 marks):
(a) n(N₂) = 14.0÷28.014 = 0.4998 mol; ÷1 = 0.4998 n(H₂) = 6.00÷2.016 = 2.976 mol; ÷3 = 0.9921N₂ has smaller quotient (0.4998 < 0.9921) → N₂ is the limiting reagent
(b) Ratio N₂:NH₃ = 1:2; n(NH₃) = 0.4998×2 = 0.9996 mol MM(NH₃) = 17.031; m(NH₃) = 0.9996×17.031 = 17.0 g (c) Ratio N₂:H₂ = 1:3; n(H₂ consumed) = 0.4998×3 = 1.499 mol n(H₂ remaining) = 2.976−1.499 = 1.477 mol; m = 1.477×2.016 = 2.98 gQ10 (4 marks):
n(NH₃) = 17.0÷17.031 = 0.9982; ÷4 = 0.2496 → NH₃ quotient n(O₂) = 48.0÷31.998 = 1.500; ÷5 = 0.3000 → O₂ quotientNH₃ has smaller quotient → NH₃ is the LR
If NH₃ is LR: ratio NH₃:NO = 4:4 = 1:1; n(NO) = 0.9982; m = 0.9982×30.006 = 29.9 g (correct) If O₂ is used incorrectly: ratio O₂:NO = 5:4; n(NO) = 1.500×(4÷5) = 1.200; m = 1.200×30.006 = 36.0 g (incorrect — too large)The second student's reasoning is wrong because using the excess reagent overestimates how much product can form. The excess reagent has "too much" relative to the LR; the reaction stops when the LR runs out, leaving excess reagent unused. The theoretical yield must be calculated only from the limiting reagent, which gives the lower (correct) answer.
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