Limiting Reactants

In many chemical reactions, the reactants are not present in exactly the proportions required by the balanced equation.

One reactant will usually be used up first.

This substance is called the limiting reactant.

The limiting reactant is important because it determines the maximum amount of product that can form.

Once the limiting reactant has been completely consumed, the reaction cannot continue—even if some of the other reactant remains.

A Simple Example

Consider:

2H₂ + O₂ → 2H₂O

The balanced equation tells us that:

2 mol H₂ react with 1 mol O₂

Suppose we have:

4 mol H₂ and 1 mol O₂

The 1 mol O₂ requires only:

2 mol H₂

But we have 4 mol H₂.

Therefore:

• all 1 mol O₂ is consumed
• only 2 mol H₂ is consumed
• 2 mol H₂ remain
• 2 mol H₂O form

In this situation:

O₂ is the limiting reactant.

H₂ is the excess reactant.

The limiting reactant controls the amount of water that can form.

Here you can explore exactly how changing the starting amounts changes which reactant limits the reaction:

Limiting and Excess Reactants

There are two important terms to distinguish.

Limiting Reactant

The limiting reactant is the reactant that is completely consumed first.

It determines the maximum amount of product that can form.

Excess Reactant

An excess reactant is present in a greater amount than required.

Some of it remains after the limiting reactant has been consumed.

Think of the limiting reactant as the ingredient that runs out first.

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A Sandwich Analogy

Suppose one sandwich requires:

2 slices of bread + 1 slice of cheese → 1 sandwich

You have:

10 slices of bread

and:

3 slices of cheese

The bread could make:

10 ÷ 2 = 5 sandwiches

The cheese could make:

3 ÷ 1 = 3 sandwiches

You can therefore make only:

3 sandwiches

Cheese is the limiting ingredient.

Bread is in excess.

After making 3 sandwiches:

6 slices of bread are used

so:

4 slices of bread remain

Chemical reactions work in much the same way.

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Why the Limiting Reactant Matters

Consider:

N₂ + 3H₂ → 2NH₃

Suppose we have:

2 mol N₂

and:

3 mol H₂

Two moles of nitrogen would require:

6 mol H₂

But only 3 mol H₂ are available.

Therefore, there is not enough hydrogen to react with all the nitrogen.

H₂ is the limiting reactant.

Once the hydrogen has been consumed, ammonia production stops.

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The Limiting Reactant Determines Product Amount

Using:

N₂ + 3H₂ → 2NH₃

Suppose:

2 mol N₂

and:

3 mol H₂

are available.

Because H₂ is limiting, calculate the product from H₂.

The mole ratio is:

3 mol H₂ : 2 mol NH₃

Therefore:

3 mol H₂ → 2 mol NH₃

The maximum amount of ammonia is:

2 mol NH₃

Even though some nitrogen remains, no more ammonia can form because there is no hydrogen left.

---

Identifying the Limiting Reactant

There are several methods.

One of the most reliable methods is:

Calculate how much product each reactant could produce.

The reactant that produces the smaller amount of product is the limiting reactant.

This method works for both mole and mass problems.

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Method: Compare Product Amounts

Consider:

2H₂ + O₂ → 2H₂O

Suppose we have:

8 mol H₂

and:

3 mol O₂

Product possible from H₂

Ratio:

2 mol H₂ : 2 mol H₂O

Therefore:

8 mol H₂ → 8 mol H₂O

Product possible from O₂

Ratio:

1 mol O₂ : 2 mol H₂O

Therefore:

3 mol O₂ → 6 mol H₂O

Compare:

H₂ could produce 8 mol H₂O

O₂ could produce 6 mol H₂O

The smaller amount is:

6 mol H₂O

Therefore:

O₂ is the limiting reactant.

The maximum product is:

6 mol H₂O

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Never Add the Product Predictions

In the previous example:

H₂ predicts:

8 mol H₂O

O₂ predicts:

6 mol H₂O

We do not calculate:

8 + 6 = 14 mol H₂O

Both calculations describe the same reaction from different reactants.

The smaller prediction determines what can actually form.

Therefore:

maximum H₂O = 6 mol

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Method: Compare Required Amounts

Another method is to determine how much of one reactant is required to react with the other.

Consider:

2H₂ + O₂ → 2H₂O

Suppose:

8 mol H₂

and:

3 mol O₂

are available.

Eight moles of H₂ require:

8 × (1 mol O₂ / 2 mol H₂)

= 4 mol O₂

But only:

3 mol O₂

are available.

Therefore, there is not enough O₂.

O₂ is limiting.

This gives the same answer as the product-comparison method.

---

Method: Divide by the Coefficient

When all reactant quantities are already in moles, there is a useful shortcut.

Divide each available mole amount by its coefficient.

For:

2H₂ + O₂ → 2H₂O

Suppose:

8 mol H₂

and:

3 mol O₂

Calculate:

H₂:

8 ÷ 2 = 4

O₂:

3 ÷ 1 = 3

The smaller value identifies the limiting reactant.

Therefore:

O₂ is limiting.

This works because it compares how many complete reaction "sets" each reactant can supply.

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Important Warning About the Shortcut

Do not simply compare the number of moles.

For:

N₂ + 3H₂ → 2NH₃

suppose we have:

2 mol N₂

and:

3 mol H₂

It might seem that N₂ is limiting because there are fewer moles of it.

But divide by the coefficients:

N₂:

2 ÷ 1 = 2

H₂:

3 ÷ 3 = 1

The smaller value is for H₂.

Therefore:

H₂ is limiting.

The reactant with fewer moles is not necessarily the limiting reactant.

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Stoichiometric Proportions

Sometimes reactants are present in exactly the required ratio.

For:

2H₂ + O₂ → 2H₂O

suppose we have:

6 mol H₂

and:

3 mol O₂

The required ratio is:

2 : 1

The available ratio is also:

6 : 3 = 2 : 1

Therefore, both reactants are consumed completely.

Neither reactant is present in excess.

The quantities are in stoichiometric proportions.

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Worked Example: Ammonia

Consider:

N₂ + 3H₂ → 2NH₃

Suppose we have:

5 mol N₂

and:

12 mol H₂

Which reactant is limiting?

Calculate product from N₂

5 mol N₂ × (2 mol NH₃ / 1 mol N₂)

= 10 mol NH₃

Calculate product from H₂

12 mol H₂ × (2 mol NH₃ / 3 mol H₂)

= 8 mol NH₃

Compare:

N₂ could produce 10 mol NH₃

H₂ could produce 8 mol NH₃

Therefore:

H₂ is the limiting reactant.

Maximum product:

8 mol NH₃

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Calculating Excess Reactant Remaining

We can also calculate how much excess reactant remains.

Using:

N₂ + 3H₂ → 2NH₃

Initial amounts:

5 mol N₂

12 mol H₂

We found:

H₂ is limiting.

How much N₂ reacts?

Ratio:

3 mol H₂ : 1 mol N₂

Therefore:

12 mol H₂ × (1 mol N₂ / 3 mol H₂)

= 4 mol N₂

Initially:

5 mol N₂

Used:

4 mol N₂

Remaining:

5 − 4 = 1 mol N₂

Therefore:

1 mol N₂ remains in excess.

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A Useful Three-Part Strategy

For most limiting-reactant problems:

Identify the limiting reactant

Determine which reactant can produce less product.

Calculate the product

Use the limiting reactant to calculate the maximum amount of product.

Calculate excess remaining

Determine how much excess reactant was consumed, then subtract:

excess remaining = excess initial − excess consumed

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Limiting Reactants with Masses

Many problems provide masses rather than moles.

In this case, first convert each reactant to moles.

The pathway becomes:

mass reactant A → mol reactant A

mass reactant B → mol reactant B

Then compare the reactants using the balanced equation.

Worked Example: Hydrogen and Oxygen by Mass

Consider:

2H₂ + O₂ → 2H₂O

Suppose:

10 g H₂

and:

64 g O₂

are available.

Use:

M(H₂) = 2 g/mol

M(O₂) = 32 g/mol

Convert H₂ to moles

10 ÷ 2 = 5 mol H₂

Convert O₂ to moles

64 ÷ 32 = 2 mol O₂

Now compare.

The reaction requires:

2 mol H₂ for every 1 mol O₂

Two moles O₂ require:

4 mol H₂

We have:

5 mol H₂

Therefore, there is more H₂ than required.

O₂ is the limiting reactant.

---

Calculate the Product

Using:

2H₂ + O₂ → 2H₂O

We have:

2 mol O₂

Ratio:

1 mol O₂ : 2 mol H₂O

Therefore:

2 mol O₂ → 4 mol H₂O

Use:

M(H₂O) = 18 g/mol

Mass:

4 × 18 = 72 g

Maximum product

72 g H₂O

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Calculate the Excess Remaining

Two moles O₂ require:

4 mol H₂

But initially we had:

5 mol H₂

Therefore:

5 − 4 = 1 mol H₂ remains

Mass:

1 × 2 = 2 g H₂

So after the reaction:

• O₂ remaining = 0 g
• H₂ remaining = 2 g
• H₂O formed = 72 g

Check conservation of mass:

Initial mass:

10 + 64 = 74 g

Final mass:

72 + 2 = 74 g

Mass is conserved.

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Worked Example: Magnesium and Oxygen

Consider:

2Mg + O₂ → 2MgO

Suppose we react:

36.45 g Mg

with:

16.0 g O₂

Use:

M(Mg) = 24.3 g/mol

M(O₂) = 32.0 g/mol

Convert Mg to moles

36.45 ÷ 24.3 = 1.50 mol Mg

Convert O₂ to moles

16.0 ÷ 32.0 = 0.500 mol O₂

The required ratio is:

2 mol Mg : 1 mol O₂

For 0.500 mol O₂, we need:

1.00 mol Mg

We have:

1.50 mol Mg

Therefore:

O₂ is limiting.

Mg is in excess.

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Calculate Magnesium Oxide Produced

Equation:

2Mg + O₂ → 2MgO

From:

0.500 mol O₂

we obtain:

1.00 mol MgO

Use:

M(MgO) = 40.3 g/mol

Therefore:

mass MgO = 1.00 × 40.3

= 40.3 g

---

Calculate Magnesium Remaining

The 0.500 mol O₂ consumes:

1.00 mol Mg

Initial Mg:

1.50 mol

Remaining:

1.50 − 1.00 = 0.50 mol Mg

Mass:

0.50 × 24.3 = 12.15 g

Check:

Initial mass:

36.45 + 16.0 = 52.45 g

Final mass:

40.3 + 12.15 = 52.45 g

Again, conservation of mass is satisfied.

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Worked Example: Aluminum and Chlorine

Consider:

2Al + 3Cl₂ → 2AlCl₃

Suppose:

27.0 g Al

react with:

71.0 g Cl₂

Use:

M(Al) = 27.0 g/mol

M(Cl₂) = 71.0 g/mol

Convert to moles

Al:

27.0 ÷ 27.0 = 1.00 mol Al

Cl₂:

71.0 ÷ 71.0 = 1.00 mol Cl₂

Notice something important:

We have equal numbers of moles, but that does not mean the reactants are present in the correct ratio.

The equation requires:

2 mol Al : 3 mol Cl₂

---

Identify the Limiting Reactant

Use the coefficient method.

Al:

1.00 ÷ 2 = 0.500

Cl₂:

1.00 ÷ 3 = 0.333

The smaller value is:

0.333

Therefore:

Cl₂ is the limiting reactant.

This is a good example of why simply comparing the number of moles does not work.

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Calculate Product Mass

Using:

2Al + 3Cl₂ → 2AlCl₃

From:

1.00 mol Cl₂

Product:

1.00 × (2/3) = 0.667 mol AlCl₃

Use:

M(AlCl₃) = 133.5 g/mol

Mass:

0.667 × 133.5 ≈ 89.0 g

Maximum product

Approximately:

89.0 g AlCl₃

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Worked Example: Iron Oxide

Consider:

4Fe + 3O₂ → 2Fe₂O₃

Suppose:

112 g Fe

and:

64 g O₂

are available.

Use:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

Convert to moles

Fe:

112 ÷ 56 = 2 mol

O₂:

64 ÷ 32 = 2 mol

Compare using coefficients

Fe:

2 ÷ 4 = 0.50

O₂:

2 ÷ 3 ≈ 0.67

The smaller value is for Fe.

Therefore:

Fe is the limiting reactant.

---

Calculate Fe₂O₃ Produced

Ratio:

4 mol Fe : 2 mol Fe₂O₃

Therefore:

2 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe)

= 1 mol Fe₂O₃

Use:

M(Fe₂O₃) = 160 g/mol

Therefore:

160 g Fe₂O₃

can theoretically form.

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Calculate Oxygen Remaining

Two moles Fe require:

2 × (3/4) = 1.5 mol O₂

Initially:

2 mol O₂

Remaining:

2 − 1.5 = 0.5 mol O₂

Mass remaining:

0.5 × 32 = 16 g O₂

Check:

Initial:

112 + 64 = 176 g

Final:

160 + 16 = 176 g

Mass is conserved.

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A General Procedure for Limiting Reactant Problems

When masses are given:

1. Balance the chemical equation.

2. Convert each reactant mass to moles.

3. Compare the mole quantities using the coefficients.

4. Identify the limiting reactant.

5. Use the limiting reactant to calculate the product.

6. If required, calculate how much excess reactant was consumed.

7. Subtract to find the amount remaining.

8. Check whether the answer is reasonable.

A useful roadmap is:

grams → moles → identify limiting reactant → product moles → product mass

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Product-Comparison Method

Another reliable strategy is to calculate how much product each reactant could theoretically produce.

Suppose:

A + 2B → 3C

Reactant A could produce:

12 mol C

Reactant B could produce:

9 mol C

The actual reaction cannot produce 12 mol because B runs out first.

Therefore:

B is limiting

and:

maximum product = 9 mol C

The rule is:

The reactant that predicts the smaller amount of product is limiting.

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Why the Limiting Reactant Controls the Reaction

Imagine a factory assembling bicycles.

Each bicycle requires:

• 1 frame
• 2 wheels

Suppose the factory has:

• 100 frames
• 160 wheels

The frames could make:

100 bicycles

The wheels could make:

160 ÷ 2 = 80 bicycles

Only:

80 bicycles

can be completed.

After that, there are still:

20 frames

but no wheels.

The wheels limit production.

Chemical reactions behave similarly.

Why Reactions Often Use Excess Reactants

In laboratory and industrial chemistry, one reactant may deliberately be supplied in excess.

Why?

An excess reactant can help ensure that the more important or expensive reactant reacts as completely as possible.

For example, if reactant A is expensive but reactant B is inexpensive, a manufacturer might use extra B.

This helps maximize the use of A.

The excess material may sometimes be:

• recovered
• recycled
• separated
• reused

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Limiting Reactants in Industrial Chemistry

Large-scale chemical production depends heavily on limiting-reactant calculations.

Chemists and engineers need to know:

• which reactant controls production
• how much product can theoretically form
• how much excess reactant remains
• how much raw material is required
• whether excess material can be recycled
• how much waste may be generated

Poor control of reactant quantities can increase both cost and waste.

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Limiting Reactants and Theoretical Yield

The theoretical yield is determined by the limiting reactant.

Suppose:

Reactant A could produce:

50 g product

Reactant B could produce:

72 g product

The reaction cannot produce 72 g because reactant A runs out first.

Therefore:

theoretical yield = 50 g

and:

A is the limiting reactant

The excess reactant cannot create additional product without more limiting reactant.

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Limiting Reactants and Conservation of Mass

Limiting-reactant calculations also demonstrate conservation of mass.

Remember that excess reactant does not disappear.

For example:

Initial reactants:

40 g A + 30 g B = 70 g

Suppose B is limiting and 10 g A remains.

If there is only one product:

product mass = 60 g

because:

60 g product + 10 g excess A = 70 g

All matter must still be accounted for.

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A More Challenging Example

Consider:

N₂ + 3H₂ → 2NH₃

Suppose:

56 g N₂

and:

12 g H₂

are available.

Use:

M(N₂) = 28 g/mol

M(H₂) = 2 g/mol

M(NH₃) = 17 g/mol

Convert to moles

N₂:

56 ÷ 28 = 2 mol N₂

H₂:

12 ÷ 2 = 6 mol H₂

Required ratio:

1 N₂ : 3 H₂

Available ratio:

2 : 6

which simplifies to:

1 : 3

Therefore, the reactants are present in exactly the required stoichiometric ratio.

Both are completely consumed.

Neither is in excess.

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Calculate the Product

From:

2 mol N₂

the equation predicts:

4 mol NH₃

Mass:

4 × 17 = 68 g NH₃

Initial mass:

56 + 12 = 68 g

Product mass:

68 g

Again:

mass is conserved

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When There Is No Excess Reactant

It is possible for reactants to be present in exactly the stoichiometric ratio.

For:

2H₂ + O₂ → 2H₂O

examples include:

2 mol H₂ + 1 mol O₂

4 mol H₂ + 2 mol O₂

10 mol H₂ + 5 mol O₂

In each case, both reactants are completely consumed.

There is no excess reactant.

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Common Mistakes

Choosing the Reactant with the Smaller Mass

The reactant with the smaller mass is not automatically limiting.

Different substances have different molar masses.

Convert masses to moles first.

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Choosing the Reactant with Fewer Moles

The reactant with fewer moles is also not automatically limiting.

The balanced equation may require different numbers of moles.

Always compare using the coefficients.

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Ignoring the Balanced Equation

Limiting-reactant calculations depend on stoichiometric ratios.

An unbalanced equation gives incorrect ratios.

Always balance first.

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Calculating Product from the Excess Reactant

Once you identify the limiting reactant, use it to calculate the maximum product.

Using the excess reactant without accounting for the limit will overestimate the product.

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Adding Two Product Predictions

If reactant A predicts 10 g product and reactant B predicts 15 g product, the answer is not:

25 g

The smaller prediction determines the theoretical yield.

---

Comparing Grams Directly

Suppose you have:

10 g A

and:

20 g B

You cannot determine which is limiting simply because A has the smaller mass.

You need:

mass → moles → mole ratio

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Forgetting About Excess Reactant

When the limiting reactant is consumed, some excess reactant may remain.

It has not disappeared.

It must be included when checking conservation of mass.

---

Subtracting Different Units

Do not subtract:

grams − moles

Convert quantities to the same unit first.

---

Using the Wrong Mole Ratio

For:

N₂ + 3H₂ → 2NH₃

the ratio H₂ : NH₃ is:

3 : 2

not:

1 : 2

Use coefficients carefully.

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Rounding Too Early

Limiting-reactant problems often contain several calculation steps.

Keep extra digits until the final answer.

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Key Terms

Limiting reactant — The reactant that is completely consumed first and determines the maximum amount of product.

Limiting reagent — Another name for the limiting reactant.

Excess reactant — A reactant present in more than the amount required to react completely with the limiting reactant.

Stoichiometric ratio — The quantitative relationship between substances given by the coefficients in a balanced equation.

Stoichiometric proportions — Reactant quantities present in exactly the ratio required by the balanced equation.

Mole ratio — A ratio between substances based on coefficients in a balanced chemical equation.

Theoretical yield — The maximum amount of product predicted from the limiting reactant.

Reactant — A starting substance in a chemical reaction.

Product — A substance formed during a chemical reaction.

Molar mass — The mass of one mole of a substance, usually expressed in g/mol.

Excess remaining — The amount of excess reactant left after the limiting reactant has been consumed.

Excess consumed — The amount of the excess reactant that actually participates in the reaction.

Stoichiometry — The quantitative study of relationships between reactants and products.

Conservation of mass — The principle that total mass is conserved during an ordinary chemical reaction.

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Key Takeaways

• The limiting reactant is the reactant that runs out first.
• The limiting reactant determines the maximum amount of product that can form.
• An excess reactant remains after the limiting reactant is consumed.
• The reactant with the smallest mass is not necessarily limiting.
• The reactant with the fewest moles is not necessarily limiting.
• Reactant amounts must be compared using the coefficients in the balanced equation.
• When masses are given, convert them to moles before comparing reactants.
• One reliable method is to calculate how much product each reactant could produce.
• The reactant producing the smaller amount of product is limiting.
• When quantities are already in moles, dividing each amount by its coefficient provides a useful shortcut.
• Product calculations must be based on the limiting reactant.
• Excess reactant remaining can be calculated using:

amount remaining = initial amount − amount consumed

• Reactants may occasionally be present in exactly the stoichiometric proportions, leaving no excess.
• The theoretical yield is determined by the limiting reactant.
• Excess reactants are often deliberately used in laboratories and industry.
• Limiting-reactant calculations are essential for predicting production, reducing waste, controlling costs, and planning chemical processes.
• Conservation of mass still applies: any unused excess reactant must be included when accounting for the final mass.

The main pathway to remember is:

BALANCE → CONVERT TO MOLES → COMPARE REACTANTS → IDENTIFY LIMITING REACTANT → CALCULATE PRODUCT → FIND EXCESS REMAINING

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Check Your Understanding

Use:

2H₂ + O₂ → 2H₂O

1. If 4 mol H₂ react with 1 mol O₂, identify the limiting reactant.

2. How many moles of H₂O can form?

3. How many moles of excess reactant remain?

4. If 6 mol H₂ react with 4 mol O₂, identify the limiting reactant.

5. Calculate the maximum amount of H₂O that can form.

6. Calculate the amount of excess reactant remaining.

7. If 10 mol H₂ react with 5 mol O₂, is either reactant in excess? Explain.

Use:

N₂ + 3H₂ → 2NH₃

8. If 4 mol N₂ react with 6 mol H₂, identify the limiting reactant.

9. Calculate the maximum number of moles of NH₃.

10. Calculate the amount of excess reactant remaining.

11. If 5 mol N₂ react with 18 mol H₂, identify the limiting reactant.

12. Calculate the maximum amount of NH₃.

13. Calculate the amount of excess reactant remaining.

14. Explain why simply comparing the number of moles of N₂ and H₂ does not reliably identify the limiting reactant.

Mass Problems

Use:

2Mg + O₂ → 2MgO

with:

M(Mg) = 24.3 g/mol

M(O₂) = 32.0 g/mol

M(MgO) = 40.3 g/mol

15. If 24.3 g Mg react with 32.0 g O₂, identify the limiting reactant.

16. Calculate the maximum mass of MgO.

17. Calculate the mass of excess reactant remaining.

18. If 48.6 g Mg react with 16.0 g O₂, identify the limiting reactant.

19. Calculate the maximum mass of MgO.

20. Calculate the mass of excess reactant remaining.

Use:

4Fe + 3O₂ → 2Fe₂O₃

with:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

M(Fe₂O₃) = 160 g/mol

21. If 112 g Fe react with 96 g O₂, identify the limiting reactant.

22. Calculate the maximum mass of Fe₂O₃.

23. Calculate the mass of excess reactant remaining.

24. If 224 g Fe react with 96 g O₂, determine whether either reactant is in excess.

25. Calculate the mass of Fe₂O₃ produced.

Challenge Problems

Use:

2Al + 3Cl₂ → 2AlCl₃

with:

M(Al) = 27.0 g/mol

M(Cl₂) = 71.0 g/mol

M(AlCl₃) = 133.5 g/mol

26. If 54.0 g Al react with 142 g Cl₂, identify the limiting reactant.

27. Calculate the maximum mass of AlCl₃.

28. Calculate the mass of excess reactant remaining.

29. If 27.0 g Al react with 106.5 g Cl₂, determine whether either reactant is in excess.

30. Calculate the mass of product.

Reasoning and Application

31. Define a limiting reactant in your own words.

32. Explain the difference between a limiting reactant and an excess reactant.

33. Explain why the limiting reactant determines the theoretical yield.

34. Why can't the reactant with the smaller mass automatically be identified as limiting?

35. Why can't the reactant with fewer moles automatically be identified as limiting?

36. Describe the product-comparison method for identifying a limiting reactant.

37. Explain why an industrial chemical process might deliberately use one reactant in excess.

38. A reaction begins with 100 g of total reactants. After the reaction, 15 g of an excess reactant remains. If there is only one product, what mass of product should be present?

39. Explain how your answer to Question 38 demonstrates conservation of mass.

40. Describe the complete procedure you would use to solve a limiting-reactant problem when the masses of two reactants are given.