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Module 2 · L13 of 20 ~45 min ⚡ +50 XP in Learn · +25 to complete

Limiting Reagents & Theoretical Yield

Most real reactions don't have reactants in perfect stoichiometric proportions, one runs out first and stops the reaction. Identifying which reactant is limiting and calculating maximum yield are among the most tested skills in HSC Chemistry.

Today's hook, Making sandwiches with 10 slices of bread and 3 fillings: filling limits how many sandwiches you can make. Chemistry works the same way, but you can't just count masses.
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Warm up and think first

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions
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Recall, your gut answer first
+5 XP warm-up

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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The five-step method, and the words for it

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Limiting Reagent Method · this lesson
core formula
Step 1: Balance the equation
Step 2: n = m ÷ MM for BOTH reactants
Step 3: Divide each n by its coefficient → compare
Step 4: Smaller value = limiting reagent (LR)
Step 5: Calculate yield using LR moles only

⚠️ Never identify the LR by comparing masses or moles alone, you must divide each by its coefficient first. A substance with fewer moles might still be in excess if its coefficient is smaller.

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What you'll master
Know

Key facts

  • Limiting reagent (LR), the reactant that runs out first
  • Excess reagent, the reactant remaining when reaction stops
  • Theoretical yield, maximum product from LR
  • LR identified by smallest n ÷ coefficient value
Understand

Concepts

  • Why the reaction stops when the LR is consumed
  • Why comparing raw moles alone is insufficient
  • Why theoretical yield is calculated from the LR only
Can do

Skills

  • Identify the LR from masses of two reactants
  • Calculate theoretical yield of any product from the LR
  • Calculate mass of excess reagent remaining after reaction
04
Key terms
Limiting reagent
The reactant that is completely consumed first; determines the maximum amount of product that can form.
Excess reagent
The reactant present in greater than stoichiometric amount; some remains unused after the reaction is complete.
Theoretical yield
The maximum mass of product calculated from the stoichiometry of the limiting reagent; assumes 100% conversion.
Identifying the limiting reagent
Convert all reactant masses to moles; divide each by its stoichiometric coefficient; the smallest result indicates the limiting reagent.
Stoichiometric calculation from LR
Use the mole ratio between limiting reagent and product to find moles of product; convert to mass using molar mass.
Excess amount remaining
n(excess remaining) = n(excess initial) − n(excess reacted); n(excess reacted) calculated from mole ratio with LR.
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Limiting and excess reagents, and theoretical yield

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Limiting and Excess Reagents
core concept · +3 XP at end

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.

Analogy:
Making sandwiches requires 2 slices of bread and 1 filling. If you have 10 slices of bread and 3 fillings, you can only make 3 sandwiches, filling is the limiting reagent. 4 slices of bread are left over (excess).
How to Identify the Limiting Reagent
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 comparison: For each reactant calculate n ÷ coefficient. The smaller value → limiting reagent. The larger value → excess reagent.

Example: 2H₂ + O₂ → 2H₂O
n(H₂) = 3 mol → 3 ÷ 2 = 1.5
n(O₂) = 2 mol → 2 ÷ 1 = 2.0
H₂ has the smaller value (1.5 < 2.0) → H₂ is the limiting reagent
Limiting reagent comparison of H₂ vs O₂ via n ÷ coefficient
Beyond the syllabus. Identifying the limiting reagent is explicit Module 2 Core and it will be examined: comparing n ÷ coefficient for each reactant is the skill being assessed on this page. Theoretical yield is the sentence after it, and the syllabus does not name it as a separate endpoint, so you will not be asked to define the term for its own sake. Learn it anyway, because it is nothing new. It is the ordinary mass-to-mass stoichiometry of Lesson 12 run from the limiting reagent's moles, and it is how the limiting reagent is usually tested in practice.

Theoretical Yield

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.

Mass of Excess Reagent Remaining
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.
Excess remaining:
n(excess consumed) = n(LR) × coeff(excess) ÷ coeff(LR)
n(excess remaining) = n(excess original) − n(excess consumed)
m(excess remaining) = n(excess remaining) × MM(excess)

The limiting reagent (LR) is identified by comparing n ÷ coefficient for each reactant, the smaller value is the LR. Theoretical yield always uses LR moles. Excess remaining = n(initial excess) − n(excess consumed), where n(excess consumed) = n(LR) × coeff(excess) ÷ coeff(LR).

Pause, copy the highlighted rule into your book before moving on.

Did you get this? True or false: in 2Na + Cl₂ → 2NaCl, if you have 0.44 mol Na and 0.28 mol Cl₂, you can identify the LR just by saying "Cl₂ has fewer moles, so it's the LR".

Quick check: 2Al + 3Cl₂ → 2AlCl₃. You have 0.40 mol Al and 0.30 mol Cl₂. Which is the limiting reagent?

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Four worked examples

Worked examples · reveal as you go

Worked example 1 · H₂ + O₂ (basic LR) +5 XP on full reveal

5.00 g of H₂ reacts with 32.0 g of O₂ to form water. Identify the limiting reagent and calculate the theoretical yield of H₂O. 2H₂ + O₂ → 2H₂O. (H = 1.008, O = 15.999)

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Balanced. Coefficients: $\mathrm{H_2} = 2$, $\mathrm{O_2} = 1$, $\mathrm{H_2O} = 2$
Extract coefficients
2
$n(\mathrm{H_2}) = 5.00 \div 2.016 = 2.480\ \mathrm{mol}$
$n(\mathrm{O_2}) = 32.0 \div 31.998 = 1.000\ \mathrm{mol}$
n for both reactants
3
$\mathrm{H_2}$: $2.480 \div 2 = 1.240$
$\mathrm{O_2}$: $1.000 \div 1 = 1.000$
O₂ smaller → O₂ is the LR
Divide by coefficient & compare
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Ratio $\mathrm{O_2}:\mathrm{H_2O} = 1:2$ → $n(\mathrm{H_2O}) = 1.000 \times 2 = 2.000\ \mathrm{mol}$
$\mathrm{MM(H_2O)} = 18.015$; $m(\mathrm{H_2O}) = 2.000 \times 18.015 =$ 36.0 g ✓
Yield from LR only
Step 1 of 4
Worked example 2 · Na + Cl₂ (non-1:1 ratio) +5 XP on full reveal

10.0 g of Na reacts with 20.0 g of Cl₂. Identify the limiting reagent and calculate the theoretical yield of NaCl. 2Na + Cl₂ → 2NaCl. (Na = 22.990, Cl = 35.453)

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Coefficients: $\mathrm{Na} = 2$, $\mathrm{Cl_2} = 1$
Balanced
2
$n(\mathrm{Na}) = 10.0 \div 22.990 = 0.4350\ \mathrm{mol}$
$n(\mathrm{Cl_2}) = 20.0 \div 70.906 = 0.2820\ \mathrm{mol}$
n for both
3
$\mathrm{Na}$: $0.4350 \div 2 = 0.2175$
$\mathrm{Cl_2}$: $0.2820 \div 1 = 0.2820$
Na smaller → Na is the LR
Note: Na had MORE moles but is still the LR
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Ratio $\mathrm{Na}:\mathrm{NaCl} = 2:2 = 1:1$; $n(\mathrm{NaCl}) = 0.4350\ \mathrm{mol}$
$\mathrm{MM(NaCl)} = 58.443$; $m(\mathrm{NaCl}) = 0.4350 \times 58.443 =$ 25.4 g ✓
Yield from LR
Step 1 of 4
Worked example 3 · Excess reagent remaining +5 XP on full reveal

From WE2: 10.0 g Na + 20.0 g Cl₂ → NaCl. Na is the LR. Calculate the mass of Cl₂ remaining after the reaction.

1
$n(\mathrm{Cl_2}\text{ consumed}) = 0.4350 \times (1 \div 2) = 0.2175\ \mathrm{mol}$
Use LR moles and ratio Na:Cl₂ = 2:1
2
$n(\mathrm{Cl_2}\text{ remaining}) = 0.2820 - 0.2175 = 0.0645\ \mathrm{mol}$
Subtract from initial
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$m(\mathrm{Cl_2}) = 0.0645 \times 70.906 =$ 4.57 g ✓
Convert to mass
Step 1 of 3
Worked example 4 · 2Al + 3Cl₂ → 2AlCl₃ (full chain) +5 XP on full reveal

10.8 g of Al reacts with 21.3 g of Cl₂. (a) Identify the LR. (b) Find theoretical yield of AlCl₃. (c) Find mass of excess reagent remaining. (Al = 26.982, Cl = 35.453)

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$2\mathrm{Al} + 3\mathrm{Cl_2} \rightarrow 2\mathrm{AlCl_3}$. Coefficients $\mathrm{Al} = 2$, $\mathrm{Cl_2} = 3$
Balanced
2
$n(\mathrm{Al}) = 10.8 \div 26.982 = 0.4003\ \mathrm{mol}$
$n(\mathrm{Cl_2}) = 21.3 \div 70.906 = 0.3004\ \mathrm{mol}$
n for both
3
$\mathrm{Al}$: $0.4003 \div 2 = 0.2002$
$\mathrm{Cl_2}$: $0.3004 \div 3 = 0.1001$
(a) Cl₂ is the LR
Smallest quotient = LR
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Ratio $\mathrm{Cl_2}:\mathrm{AlCl_3} = 3:2$; $n(\mathrm{AlCl_3}) = 0.3004 \times (2 \div 3) = 0.2003\ \mathrm{mol}$
$\mathrm{MM(AlCl_3)} = 26.982 + 3(35.453) = 133.34$; $m = 0.2003 \times 133.34 =$ (b) 26.7 g
Yield from LR
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$n(\mathrm{Al}\text{ consumed}) = 0.3004 \times (2 \div 3) = 0.2003\ \mathrm{mol}$
$n(\mathrm{Al}\text{ remaining}) = 0.4003 - 0.2003 = 0.2000\ \mathrm{mol}$
$m(\mathrm{Al}\text{ remaining}) = 0.2000 \times 26.982 =$ (c) 5.40 g
Excess remaining
Step 1 of 5

Match each term to its definition.

  • Limiting reagent
  • Excess reagent
  • Theoretical yield
  • Mole ratio
  • The maximum mass (or moles) of product calculable from the limiting reagent and a 100% conversion.
  • The reactant that runs out first and therefore caps the amount of product formed.
  • The ratio of stoichiometric coefficients used to convert between moles of two species in the balanced equation.
  • The reactant present in more moles than needed by the mole ratio; some remains unreacted at the end.
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Put the method in order

Sort the steps given the masses of two reactants, find the theoretical yield of the product. Click two steps to swap them, then check the order.

  • For each reactant, calculate n ÷ coefficient, the smaller value is the limiting reagent
  • Write the balanced equation and note the coefficients
  • m(product) = n(product) × MM(product), the theoretical yield
  • Convert each reactant mass to moles: n = m ÷ MM
  • n(product) = n(limiting reagent) × (coefficient of product ÷ coefficient of limiting reagent)
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The three traps that cost marks

Common errors · the 3 traps that cost marks

1

Picking the LR based on smaller mass or smaller moles alone

In 2Na + Cl₂ → 2NaCl with 0.44 mol Na and 0.28 mol Cl₂, a student might say "Cl₂ is the LR because it has fewer moles." But you must divide by coefficients: Na = 0.44 ÷ 2 = 0.22, Cl₂ = 0.28 ÷ 1 = 0.28. Na is actually the LR.

Fix: Always perform n ÷ coefficient for every reactant before comparing. Never compare raw moles or masses.

2

Using the excess reagent's moles to calculate yield

Once the LR is identified, all product yield calculations must use the LR's moles. Using the excess reagent's moles gives a larger, incorrect theoretical yield, particularly dangerous because the number seems plausible.

Fix: After identifying the LR, cross out the excess reagent's moles. Proceed from LR moles only for all yield calculations.

3

Forgetting to subtract consumed moles when finding excess remaining

The mass of excess reagent remaining requires two steps: (1) calculate how much was consumed using the LR and the ratio, (2) subtract from the original amount. Reporting just the original mass of the excess reagent is wrong.

Fix: n(excess remaining) = n(excess original) − n(excess consumed). Always calculate n(consumed) first using LR moles and the equation ratio.

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Quick-fire drill, then revisit your thinking

Quick-fire practice · 5 reps +2 XP per reveal

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)

5

Following Q4: calculate the theoretical mass of NH₃ produced from 5.60 g N₂ and 1.50 g H₂. (MM(NH₃) = 17.031)

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Revisit your thinking

Earlier you were asked: How would you decide which of two reactants runs out first?

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.

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Interactive Tool, Stoichiometry Calculator Open fullscreen ↗
Use the Stoichiometry Calculator. How many moles are in 44 g of CO₂ (molar mass = 44 g/mol)?
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Practice questions

01
Multiple choice
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02
Short answer
ApplyLower-order5 marks

Q1. 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)

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AnalyseMid-order6 marks

Q2. 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)

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EvaluateHigher-order3 marks

Q3. 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.

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AnalyseMid-order6 marks

Q4. 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)

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EvaluateHigher-order4 marks

Q5. 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 LR 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)

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📖 Comprehensive answers (click to reveal)

Multiple choice, drill bank

MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.

Short answer model answers

Q1 (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 LR
(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 g

Q2 (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; m(CO₂) = 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 = 0.005×31.998 = 0.16 g

Q3 (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.

Q4 (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 = 0.9921
N₂ has smaller quotient (0.4998 < 0.9921) → N₂ is the LR
(b) Ratio N₂:NH₃ = 1:2; n(NH₃) = 0.4998×2 = 0.9996 mol; MM(NH₃) = 17.031; m = 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 g

Q5 (4 marks):

n(NH₃) = 17.0÷17.031 = 0.9982; ÷4 = 0.2496
n(O₂) = 48.0÷31.998 = 1.500; ÷5 = 0.3000
NH₃ smaller → 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 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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