Limiting Reactant vs. Excess Reactant

These two concepts are closely connected.

Limiting Reactant

The reactant that:

• is completely consumed first
• stops the reaction when it runs out
• determines the maximum amount of product

Excess Reactant

The reactant that:

• is supplied in more than the required amount
• is not completely consumed
• remains after the reaction stops

If one reactant is limiting, another reactant is usually in excess.

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Why Does an Excess Reactant Remain?

Chemical reactions occur according to specific particle ratios.

Consider:

N₂ + 3H₂ → 2NH₃

Every:

1 mol N₂

requires:

3 mol H₂

Suppose we have:

2 mol N₂

and:

9 mol H₂

The 2 mol N₂ require:

6 mol H₂

But:

9 mol H₂

are available.

Only 6 mol H₂ can react because all the nitrogen is then gone.

Therefore:

9 − 6 = 3 mol H₂

remain.

Hydrogen is the excess reactant.

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The Basic Excess Reactant Calculation

The central calculation is:

amount remaining = amount initially present − amount consumed

The difficult part is usually determining the amount consumed.

To find it:

1. Identify the limiting reactant.
2. Use the limiting reactant and the mole ratio to determine how much excess reactant reacts.
3. Subtract that amount from the initial amount.

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A Useful Roadmap

For excess-reactant problems:

BALANCE THE EQUATION

↓

CONVERT REACTANTS TO MOLES

↓

IDENTIFY THE LIMITING REACTANT

↓

CALCULATE EXCESS REACTANT CONSUMED

↓

INITIAL EXCESS − CONSUMED EXCESS

↓

EXCESS REACTANT REMAINING

This is closely related to limiting-reactant calculations, but the final goal is different.

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Worked Example: Hydrogen and Oxygen

Consider:

2H₂ + O₂ → 2H₂O

Suppose:

10 mol H₂

and:

3 mol O₂

are available.

Identify the limiting reactant

Three moles of O₂ require:

3 × 2 = 6 mol H₂

We have:

10 mol H₂

Therefore:

O₂ is limiting

and:

H₂ is in excess

Determine how much H₂ reacts

From the equation:

1 mol O₂ requires 2 mol H₂

Therefore:

3 mol O₂ require 6 mol H₂

Calculate the amount remaining

Initial H₂:

10 mol

Consumed:

6 mol

Remaining:

10 − 6 = 4 mol

Answer

4 mol H₂ remain after the reaction.

---

Calculate the Product Too

Using the same reaction:

2H₂ + O₂ → 2H₂O

and:

10 mol H₂ + 3 mol O₂

we determined that O₂ is limiting.

The ratio is:

1 mol O₂ : 2 mol H₂O

Therefore:

3 mol O₂ → 6 mol H₂O

At the end:

• H₂O formed = 6 mol
• O₂ remaining = 0 mol
• H₂ remaining = 4 mol

This gives us a complete picture of the reaction.

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Worked Example: Ammonia

Consider:

N₂ + 3H₂ → 2NH₃

Suppose:

5 mol N₂

and:

12 mol H₂

are available.

Identify the limiting reactant

Five moles N₂ would require:

15 mol H₂

But only:

12 mol H₂

are available.

Therefore:

H₂ is limiting

and:

N₂ is in excess

Calculate N₂ consumed

Ratio:

3 mol H₂ : 1 mol N₂

Therefore:

12 mol H₂ × (1 mol N₂ / 3 mol H₂)

= 4 mol N₂

Calculate N₂ remaining

Initial:

5 mol N₂

Consumed:

4 mol N₂

Remaining:

5 − 4 = 1 mol N₂

Answer

1 mol N₂ remains in excess.

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Worked Example: Another Ammonia Problem

Suppose:

4 mol N₂

and:

9 mol H₂

are available.

Equation:

N₂ + 3H₂ → 2NH₃

Nine moles H₂ require:

9 × (1/3) = 3 mol N₂

We have:

4 mol N₂

Therefore:

H₂ is limiting

and:

N₂ is in excess

N₂ remaining:

4 − 3 = 1 mol N₂

Product:

9 mol H₂ × (2 mol NH₃ / 3 mol H₂)

= 6 mol NH₃

At the end:

• H₂ = 0 mol
• N₂ = 1 mol
• NH₃ = 6 mol

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When Masses Are Given

Excess-reactant problems often provide masses rather than moles.

Because balanced equations describe mole ratios, first convert the reactants to moles.

The pathway becomes:

grams → moles → identify limiting/excess reactants → calculate excess consumed → calculate excess remaining

Worked Example: Magnesium and Oxygen

Consider:

2Mg + O₂ → 2MgO

Suppose:

36.45 g Mg

react with:

16.0 g O₂

Use:

M(Mg) = 24.3 g/mol

M(O₂) = 32.0 g/mol

Convert magnesium to moles

36.45 ÷ 24.3 = 1.50 mol Mg

Convert oxygen to moles

16.0 ÷ 32.0 = 0.500 mol O₂

Identify the limiting reactant

The equation requires:

2 mol Mg : 1 mol O₂

Therefore:

0.500 mol O₂

requires:

1.00 mol Mg

We have:

1.50 mol Mg

Therefore:

O₂ is limiting

and:

Mg is in excess

---

Calculate the Excess Magnesium Remaining

Initial Mg:

1.50 mol

Mg consumed:

1.00 mol

Therefore:

1.50 − 1.00 = 0.50 mol Mg

Convert to mass:

m = nM

m = 0.50 × 24.3

= 12.15 g

Answer

12.15 g Mg remain after the reaction.

---

Check the Entire Reaction

The limiting O₂ produces MgO.

From:

2Mg + O₂ → 2MgO

0.500 mol O₂ → 1.00 mol MgO

Use:

M(MgO) = 40.3 g/mol

Therefore:

40.3 g MgO

are produced.

Initial mass:

36.45 + 16.0 = 52.45 g

Final mass:

40.3 + 12.15 = 52.45 g

Therefore:

initial mass = final mass

The leftover excess reactant must be included when checking conservation of mass.

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Why the Excess Reactant Cannot Keep Reacting

Suppose magnesium remains after all the oxygen has been consumed.

You might ask:

Why doesn't the remaining magnesium continue reacting?

Because the reaction requires oxygen.

Once there are no O₂ molecules left, the remaining Mg atoms have nothing to react with.

The reaction stops even though magnesium is still present.

Adding more oxygen would allow the reaction to continue.

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Worked Example: Iron and Oxygen

Consider:

4Fe + 3O₂ → 2Fe₂O₃

Suppose:

112 g Fe

react with:

64 g O₂

Use:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

Convert to moles

Fe:

112 ÷ 56 = 2.00 mol

O₂:

64 ÷ 32 = 2.00 mol

Equal numbers of moles do not mean equal stoichiometric quantities.

The equation requires:

4 mol Fe : 3 mol O₂

---

Identify the Excess Reactant

Two moles Fe require:

2 × (3/4) = 1.50 mol O₂

But:

2.00 mol O₂

are available.

Therefore:

Fe is limiting

and:

O₂ is in excess

---

Calculate Oxygen Remaining

Initial O₂:

2.00 mol

Consumed:

1.50 mol

Remaining:

2.00 − 1.50 = 0.50 mol O₂

Convert to mass:

0.50 × 32 = 16 g

Answer

16 g O₂ remain.

---

Check the Product

Two moles Fe produce:

1 mol Fe₂O₃

Use:

M(Fe₂O₃) = 160 g/mol

Product:

160 g Fe₂O₃

Initial mass:

112 + 64 = 176 g

Final mass:

160 + 16 = 176 g

Again:

mass is conserved

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Worked Example: Aluminum and Chlorine

Consider:

2Al + 3Cl₂ → 2AlCl₃

Suppose:

54.0 g Al

react with:

142 g Cl₂

Use:

M(Al) = 27.0 g/mol

M(Cl₂) = 71.0 g/mol

Convert to moles

Al:

54.0 ÷ 27.0 = 2.00 mol Al

Cl₂:

142 ÷ 71.0 = 2.00 mol Cl₂

Required ratio:

2 mol Al : 3 mol Cl₂

Two moles Cl₂ require:

2 × (2/3) = 1.33 mol Al

We have:

2.00 mol Al

Therefore:

Cl₂ is limiting

and:

Al is in excess

---

Calculate Aluminum Remaining

Al consumed:

2.00 mol Cl₂ × (2 mol Al / 3 mol Cl₂)

= 1.33 mol Al

Initial Al:

2.00 mol

Remaining:

2.00 − 1.33 = 0.67 mol Al

Convert to mass:

0.67 × 27.0 ≈ 18.0 g Al

Answer

Approximately:

18.0 g Al remain.

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Using Product Amount to Find Excess Consumed

Sometimes you already know how much product formed.

Consider:

2H₂ + O₂ → 2H₂O

Suppose the reaction produces:

8 mol H₂O

How much O₂ was consumed?

Ratio:

1 mol O₂ : 2 mol H₂O

Therefore:

8 mol H₂O × (1 mol O₂ / 2 mol H₂O)

= 4 mol O₂

If initially there were:

6 mol O₂

then:

6 − 4 = 2 mol O₂

remain.

This is another way to calculate leftover reactant.

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Calculating Percent Excess

In more advanced stoichiometry, chemists may describe how much extra reactant has been supplied using percent excess.

First determine how much reactant is actually required.

Then:

excess amount = actual amount − required amount

and:

percent excess = (excess amount / required amount) × 100%

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Worked Example: Percent Excess

Suppose a reaction requires:

20 g of reactant B

but:

25 g

are supplied.

Excess:

25 − 20 = 5 g

Percent excess:

(5 / 20) × 100% = 25%

Therefore:

B was supplied at 25% excess.

This does not mean that 25% of the original amount necessarily remains in every situation; it describes the extra amount relative to the stoichiometric requirement.

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Why Use an Excess Reactant?

Using an excess reactant may sound wasteful, but it can be useful.

An excess reactant may help:

• ensure the limiting reactant reacts completely
• increase conversion of an expensive reactant
• improve production efficiency
• drive some reactions toward greater product formation
• compensate for practical losses
• maintain desired reaction conditions

The choice of which substance to use in excess can be economically important.

Excess Reactants in Industry

Suppose reactant A is very expensive and reactant B is inexpensive.

A manufacturer may deliberately use excess B.

This helps ensure that as much of A as possible reacts.

Afterward, unused B may sometimes be:

• separated
• purified
• recycled
• returned to the reactor

This can reduce both costs and waste.

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Excess Oxygen in Combustion

Combustion provides a familiar example.

For complete methane combustion:

CH₄ + 2O₂ → CO₂ + 2H₂O

The stoichiometric ratio requires:

1 mol CH₄ : 2 mol O₂

In practical combustion systems, oxygen may be supplied in excess to help ensure more complete combustion of the fuel.

Insufficient oxygen can contribute to incomplete combustion and the formation of products such as carbon monoxide.

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Worked Combustion Example

Suppose:

2 mol CH₄

react with:

6 mol O₂

Equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

Two moles CH₄ require:

4 mol O₂

Available:

6 mol O₂

Therefore:

CH₄ is limiting

and:

O₂ is in excess

O₂ remaining:

6 − 4 = 2 mol O₂

Products:

2 mol CO₂

and:

4 mol H₂O

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When There Is No Excess Reactant

Not every reaction mixture has an excess reactant.

For:

2H₂ + O₂ → 2H₂O

suppose:

4 mol H₂

and:

2 mol O₂

are available.

The ratio is exactly:

2 : 1

Both reactants are completely consumed.

Therefore:

H₂ remaining = 0

O₂ remaining = 0

The reactants were present in stoichiometric proportions.

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Excess Reactants and Conservation of Mass

Excess reactants are especially important when accounting for mass.

Suppose:

70 g

of reactants are initially present.

After the reaction:

12 g

of an excess reactant remain.

If there is only one product:

product mass = 70 − 12

= 58 g

The excess reactant remains part of the system.

It cannot simply be ignored.

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A Complete Mass Balance

Suppose a reaction begins with:

25 g A

and:

40 g B

Total initial mass:

65 g

Suppose A is limiting and:

15 g B

remain afterward.

Mass consumed:

25 g A + 25 g B = 50 g

If there is one product:

product mass = 50 g

Final mass:

50 g product + 15 g B = 65 g

Therefore:

initial mass = final mass

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Practical Laboratory Example

Imagine mixing two solutions to form a precipitate.

If one dissolved reactant is supplied in excess:

• all of the limiting reactant may be consumed
• the solid product forms
• some excess reactant remains dissolved in the solution

The excess reactant has not disappeared simply because it cannot be seen.

It may remain as dissolved ions in the solution.

This is important when interpreting laboratory results.

Choosing Which Reactant Should Be in Excess

Chemists may consider several factors:

• cost
• availability
• safety
• toxicity
• ease of separation
• environmental impact
• ability to recycle unused material
• desired reaction efficiency

For example, it may make sense to use an inexpensive and easily removed substance in excess rather than an expensive or hazardous one.

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A Full Problem-Solving Example

Consider:

2Na + Cl₂ → 2NaCl

Suppose:

69 g Na

react with:

71 g Cl₂

Use:

M(Na) = 23 g/mol

M(Cl₂) = 71 g/mol

M(NaCl) = 58.5 g/mol

Convert to moles

Na:

69 ÷ 23 = 3 mol Na

Cl₂:

71 ÷ 71 = 1 mol Cl₂

Determine the limiting reactant

One mole Cl₂ requires:

2 mol Na

We have:

3 mol Na

Therefore:

Cl₂ is limiting

and:

Na is in excess

Calculate Na consumed

1 mol Cl₂ × (2 mol Na / 1 mol Cl₂)

= 2 mol Na

Calculate Na remaining

Initial:

3 mol Na

Consumed:

2 mol Na

Remaining:

1 mol Na

Mass remaining:

1 × 23 = 23 g Na

Calculate product

One mole Cl₂ produces:

2 mol NaCl

Mass:

2 × 58.5 = 117 g NaCl

Check mass conservation

Initial:

69 + 71 = 140 g

Final:

117 + 23 = 140 g

Everything is accounted for.

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Common Mistakes

Confusing Excess with Limiting

The limiting reactant runs out.

The excess reactant remains.

---

Assuming the Larger Mass Is Excess

A substance is not excess simply because more grams are present.

Molar masses and stoichiometric ratios must be considered.

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Assuming the Larger Number of Moles Is Excess

The balanced equation may require different mole quantities.

For:

N₂ + 3H₂ → 2NH₃

having more H₂ moles than N₂ does not automatically mean H₂ is in excess.

Three times as much H₂ is required.

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Subtracting the Limiting Reactant from the Excess Reactant

Do not calculate:

initial excess − limiting amount

unless the mole ratio happens to be 1 : 1.

First use the balanced equation to determine the amount of excess reactant consumed.

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Subtracting Different Units

Do not calculate something such as:

10 g − 0.2 mol

Both quantities must be expressed in compatible units.

---

Forgetting to Convert Back to Grams

If the question asks for the mass remaining, convert leftover moles to mass.

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Using Initial Excess to Calculate Product

The product is controlled by the limiting reactant, not by the total amount of excess reactant supplied.

---

Forgetting Leftover Material in Mass Conservation

If excess reactant remains, it must be included in the final mass.

---

Rounding Too Early

Keep additional digits during intermediate calculations and round at the end.

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Key Terms

Excess reactant — A reactant present in more than the stoichiometric amount required.

Limiting reactant — The reactant consumed first, which determines the maximum amount of product.

Leftover reactant — The portion of an excess reactant remaining after the reaction stops.

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

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

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

Stoichiometric proportions — Reactants present in exactly the required mole ratio.

Mole ratio — A relationship between quantities of substances based on balanced-equation coefficients.

Percent excess — The amount supplied beyond the stoichiometric requirement, expressed as a percentage of the required amount.

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

Mass balance — Accounting for all mass entering, leaving, reacting, and remaining in a chemical system.

Conservation of mass — The principle that total mass remains constant during an ordinary chemical reaction.

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Key Takeaways

• An excess reactant is present in more than the amount required by the balanced equation.
• The excess reactant is not completely consumed.
• The limiting reactant runs out first.
• The limiting reactant determines how much product forms.
• The excess reactant determines how much material may remain afterward.
• Limiting and excess reactants must be identified using stoichiometric ratios.
• Neither mass nor number of moles alone reliably identifies the excess reactant.
• When masses are given, convert them to moles before comparing reactants.
• To find leftover reactant:

amount remaining = amount initially present − amount consumed

• Use the limiting reactant to calculate how much of the excess reactant is consumed.
• If the question asks for leftover mass, convert the remaining moles back to grams.
• Sometimes reactants are present in exact stoichiometric proportions and neither remains in excess.
• Excess reactants may be deliberately used in laboratories and industrial processes.
• Excess reactants can help ensure that a more valuable reactant is consumed as completely as possible.
• Unused excess material may sometimes be recovered and recycled.
• Leftover reactant must be included when checking conservation of mass.

The main pathway is:

BALANCE → CONVERT TO MOLES → IDENTIFY LIMITING REACTANT → IDENTIFY EXCESS REACTANT → CALCULATE EXCESS CONSUMED → SUBTRACT → FIND EXCESS REMAINING

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Check Your Understanding

Use:

2H₂ + O₂ → 2H₂O

1. If 8 mol H₂ react with 3 mol O₂, identify the excess reactant.

2. How many moles of the excess reactant are consumed?

3. How many moles of the excess reactant remain?

4. How many moles of H₂O form?

5. If 6 mol H₂ react with 3 mol O₂, is either reactant in excess? Explain.

Use:

N₂ + 3H₂ → 2NH₃

6. If 4 mol N₂ react with 9 mol H₂, identify the excess reactant.

7. Calculate the amount of N₂ consumed.

8. Calculate the amount of N₂ remaining.

9. Calculate the amount of NH₃ produced.

10. If 3 mol N₂ react with 12 mol H₂, calculate the amount of excess reactant remaining.

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

11. If 48.6 g Mg react with 16.0 g O₂, identify the excess reactant.

12. Calculate the mass of excess reactant consumed.

13. Calculate the mass of excess reactant remaining.

14. Calculate the mass of MgO formed.

15. Show that the initial and final masses are equal.

Use:

4Fe + 3O₂ → 2Fe₂O₃

with:

M(Fe) = 56 g/mol

M(O₂) = 32 g/mol

M(Fe₂O₃) = 160 g/mol

16. If 112 g Fe react with 96 g O₂, identify the excess reactant.

17. Calculate the amount of excess reactant consumed.

18. Calculate the mass of excess reactant remaining.

19. Calculate the mass of Fe₂O₃ produced.

20. Check your answer using conservation of mass.

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

21. If 81.0 g Al react with 213 g Cl₂, determine whether either reactant is in excess.

22. If 81.0 g Al react with 142 g Cl₂, identify the excess reactant.

23. Calculate the mass of excess reactant remaining in Question 22.

24. Calculate the mass of AlCl₃ produced.

25. Demonstrate that mass is conserved.

Use:

CH₄ + 2O₂ → CO₂ + 2H₂O

with:

M(CH₄) = 16 g/mol

M(O₂) = 32 g/mol

26. If 32 g CH₄ react with 160 g O₂, identify the excess reactant.

27. Calculate the mass of excess reactant consumed.

28. Calculate the mass of excess reactant remaining.

29. Calculate the moles of CO₂ and H₂O produced.

30. Explain why the reaction stops even though one reactant remains.

Reasoning and Application

31. Define an excess reactant in your own words.

32. Explain the relationship between limiting and excess reactants.

33. Why can't the reactant with the larger mass automatically be identified as excess?

34. Why must the balanced equation be used when calculating leftover reactant?

35. A reaction begins with 80 g of total reactants and leaves 12 g of excess reactant. If only one product forms, calculate the mass of product.

36. Explain how Question 35 demonstrates conservation of mass.

37. A reaction requires 40 g of reactant B, but 50 g are supplied. Calculate the mass supplied in excess.

38. Calculate the percent excess in Question 37.

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

40. Describe the complete procedure for determining the mass of an excess reactant remaining when the initial masses of two reactants are known.