Complete Guide to Limiting Reagent and Excess Reagent
In stoichiometry, when reactants are not present in the exact ratio required by a balanced chemical equation, one reactant will be completely consumed before the others. This guide explains everything about limiting reagent and excess reagent.
1. Key Definitions
- Limiting Reagent (Limiting Reactant): The reactant that is completely used up first in a chemical reaction. It determines the maximum amount of product that can be formed.
- Excess Reagent (Excess Reactant): The reactant(s) that remain after the reaction is complete because they were present in a greater amount than needed.
- Theoretical Yield: The maximum amount of product that can be produced based on the limiting reagent.
- Actual Yield: The real amount of product obtained (usually less than theoretical yield).
- Percent Yield: (Actual Yield / Theoretical Yield) × 100%
If you have 10 slices of bread and 3 slices of cheese → you can only make 3 sandwiches.
Cheese is the limiting reagent, bread is in excess (4 slices left over).
400 wheels can make 100 cars
125 steering wheels can make 125 cars
So, wheels are limiting reagent and steering wheels are excess reagent
2. Step-by-Step Method to Find Limiting Reagent
- Write and balance the chemical equation.
- Convert given amounts of reactants to moles. (if given in grams, use molar mass)
- Divide each mole value by its stoichiometric coefficient from the balanced equation.
- The reactant with the smallest value after division is the limiting reagent.
- The others are excess reagents.
The reactant that produces the least amount of product (or has the smallest "moles / coefficient") is limiting.
3. Example 1: Classic Problem
Question:
When 28.0 g of nitrogen gas reacts with 12.0 g of hydrogen gas to form ammonia, which is the limiting reagent?
N₂(g) + 3H₂(g) → 2NH₃(g)
Solution:
| Reactant | Given Mass | Molar Mass | Moles | Moles ÷ Coefficient |
|---|---|---|---|---|
| N₂ | 28.0 g | 28.0 g/mol | 1.00 mol | 1.00 ÷ 1 = 1.00 |
| H₂ | 12.0 g | 2.02 g/mol | 5.94 mol | 5.94 ÷ 3 = 1.98 |
→ N₂ gives the smaller number → N₂ is the limiting reagent.
H₂ is in excess.
4. Example 2: Finding Amount of Excess Reagent
Question:
50.0 g of calcium carbonate (CaCO₃) reacts with 36.5 g of HCl. How much HCl remains after the reaction?
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
Step 1: Find moles
- CaCO₃: 50.0 g ÷ 100.1 g/mol = 0.4995 mol
- HCl: 36.5 g ÷ 36.5 g/mol = 1.000 mol
Step 2: Divide by coefficients
- CaCO₃: 0.4995 ÷ 1 = 0.4995
- HCl: 1.000 ÷ 2 = 0.500
→ CaCO₃ is limiting (smaller value).
Step 3: How much HCl is actually used?
From equation: 1 mol CaCO₃ requires 2 mol HCl
So 0.4995 mol CaCO₃ × 2 = 0.999 mol HCl used
Excess HCl = 1.000 mol – 0.999 mol = 0.001 mol ≈ 0.04 g (negligible, but shows method)
5. Alternative Method: Calculate Product from Each Reactant
Another reliable way:
- Assume reactant A is limiting → calculate how much product it makes.
- Assume reactant B is limiting → calculate how much product it makes.
- The reactant that produces less product is the limiting reagent.
2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu
5.40 g Al and 31.9 g CuSO₄ are mixed.
From Al: 5.40 g Al × (1 mol/27) × (3 mol Cu/2 mol Al) = 0.300 mol Cu
From CuSO₄: 31.9 g CuSO₄ × (1 mol/159.6) × (3 mol Cu/3 mol CuSO₄) = 0.600 mol Cu
→ Al produces less Cu → Al is limiting.
6. Summary Table: How to Identify
| Method | How to Use | Best When |
|---|---|---|
| Moles ÷ Coefficient | Smallest value = limiting | Fastest for most problems |
| Calculate product from each | Reactant giving least product = limiting | Very reliable, good for double-checking |
| Convert everything to one reactant | See which runs out first | Useful for excess calculations |
7. Common Mistakes to Avoid
- Forgetting to balance the equation first.
- Using mass instead of moles.
- Forgetting to divide by stoichiometric coefficients.
- Mixing up which reactant produces less product.
Key Takeaway
The limiting reagent is the one that runs out first — it controls how much product you can make.
Always work in moles and respect the stoichiometric ratios.
8. Percent Yield Calculations
Percent yield measures how efficient a real reaction was compared to the ideal (theoretical) result.
- Theoretical Yield: Maximum product possible, calculated from the limiting reagent using stoichiometry.
- Actual Yield: Amount of product actually obtained in the lab (always ≤ theoretical).
Values above 100% usually mean impure product or calculation error!
Step-by-Step Procedure
- Identify the limiting reagent.
- Calculate the theoretical yield of the desired product (in grams).
- Use the given/measured actual yield.
- Apply the percent yield formula.
Example 1 – Ammonia Synthesis
Reaction: N₂ + 3H₂ → 2NH₃
28.0 g N₂ reacts completely with excess H₂.
In the lab, 30.0 g NH₃ is obtained.
Step 1: Theoretical yield
28.0 g N₂ × (1 mol / 28.0 g) × (2 mol NH₃ / 1 mol N₂) × (17.0 g/mol) = 34.0 g NH₃
Step 2: Percent yield
(30.0 g / 34.0 g) × 100% = 88.2%
Example 2 – Combustion of Propane
Reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
44.0 g propane burns in excess oxygen and produces 102.0 g CO₂.
Theoretical CO₂ = 44.0 g C₃H₈ × (3 mol CO₂ / 1 mol C₃H₈) × 44.0 g/mol = 132.0 g
Percent yield = (102.0 / 132.0) × 100% = 77.3%
Example 3 – With Limiting Reagent Identification
Reaction: 2Al + 3Cl₂ → 2AlCl₃
5.40 g Al + 15.0 g Cl₂ → actual yield of AlCl₃ = 12.0 g
First, find limiting reagent:
Al: 5.40 / 27.0 = 0.200 mol ÷ 2 = 0.100
Cl₂: 15.0 / 71.0 ≈ 0.211 mol ÷ 3 ≈ 0.070 → Cl₂ is limiting
Theoretical AlCl₃ = 0.211 mol Cl₂ × (2 mol AlCl₃ / 3 mol Cl₂) × 133.3 g/mol ≈ 18.8 g
Percent yield = (12.0 / 18.8) × 100% ≈ 63.8%
Typical Percent Yield Ranges
| Situation | Typical Percent Yield |
|---|---|
| High-school lab | 60–90% |
| College organic synthesis | 40–80% |
| Industrial chemical process | 90–99+% |
| First-time or difficult reaction | <50% |
100% → Usually impossible in real life (but great on paper)
Quick Summary
- Always base theoretical yield on the limiting reagent.
- Percent yield tells you how much product was lost to side reactions, transfers, etc.
- Practice: limiting reagent → theoretical yield → percent yield is the full stoichiometry sequence!
9. Excess Reagent Calculations
After identifying the limiting reagent, you can calculate exactly how much of the excess reagent is left over at the end of the reaction.
Step-by-Step Method (Always in Moles!)
- Identify the limiting reagent.
- Use the limiting reagent and the balanced equation to calculate how many moles of the excess reagent are actually consumed.
- Subtract consumed moles from initial moles → remaining moles.
- Convert remaining moles back to grams (if required).
Example 1 – Hydrogen + Oxygen
2H₂ + O₂ → 2H₂O
8.0 g H₂ reacts with 20.0 g O₂.
Moles H₂ = 8.0 / 2.02 ≈ 3.96 mol
Moles O₂ = 20.0 / 32.0 = 0.625 mol
Divide by coefficients:
H₂ → 3.96 ÷ 2 = 1.98
O₂ → 0.625 ÷ 1 = 0.625 → O₂ is limiting
H₂ consumed = 0.625 mol O₂ × (2 mol H₂ / 1 mol O₂) = 1.25 mol
H₂ left = 3.96 – 1.25 = 2.71 mol → 2.71 × 2.02 ≈ 5.5 g H₂ left over
Example 2 – Classic CaCO₃ + HCl
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
25.0 g CaCO₃ reacts with 29.2 g HCl.
Moles CaCO₃ = 25.0 / 100.1 ≈ 0.250 mol → limiting
HCl consumed = 0.250 × 2 = 0.500 mol
HCl initial = 29.2 / 36.5 ≈ 0.800 mol
HCl left = 0.800 – 0.500 = 0.300 mol → 0.300 × 36.5 = 11.0 g HCl excess
Example 3 – Multiple Excess Reagents
4NH₃ + 5O₂ → 4NO + 6H₂O
68.0 g NH₃ and 160.0 g O₂ are mixed.
NH₃: 68.0 / 17.0 = 4.00 mol ÷ 4 = 1.00
O₂: 160.0 / 32.0 = 5.00 mol ÷ 5 = 1.00 → stoichiometric! → no excess
If you had 170.0 g O₂ instead → O₂ moles = 5.3125 → 5.3125 ÷ 5 = 1.0625
NH₃ is now limiting → O₂ excess = 5.3125 – (4 × 5) = 0.3125 mol ≈ 10.0 g O₂ left
Quick Formula Shortcut (Once Limiting is Known)
| Situation | Excess Mass Formula |
|---|---|
| A is limiting, find excess B | Excess B (g) = Initial B (g) − [moles A × (coeff. B / coeff. A) × molar mass B] |
Common Mistakes to Avoid
- Subtracting masses directly (wrong!)
- Forgetting the stoichiometric ratio (2:1, 3:1, etc.)
- Using the wrong reagent as limiting
- Not converting back to grams when asked for “mass left”
1. Find limiting reagent
2. Calculate how much excess reagent it consumes
3. Initial − consumed = leftover
10. Practice Problems (with Detailed Solutions)
Click on each problem to reveal the step-by-step solution.
Problem 1 (Easy): 2Na + S → Na₂S
9.2 g of sodium is reacted with 8.0 g of sulfur to produce sodium sulfide. Which is the limiting reactant?
Moles Na = 9.2 / 23 ≈ 0.40 mol → divide by coeff. 2 → 0.20
Moles S = 8.0 / 32.0 = 0.25 mol → divide by coeff. 1 → 0.25
Na has the smaller value → Na is limiting.
Excess S = (0.25 – 0.20) mol = 0.05 mol ≈ 1.6 g left.
Problem 2 (Easy): 2H₂ + O₂ → 2H₂O
6.0 g of H₂ reacts with 40.0 g of O₂. Find the limiting reagent.
Moles H₂ = 6.0 / 2.02 ≈ 2.97 mol → divide by coeff. 2 → 1.485
Moles O₂ = 40.0 / 32.0 = 1.25 mol → divide by coeff. 1 → 1.25
O₂ has the smaller value → O₂ is limiting.
Excess H₂ = (2.97 – 2 × 1.25) mol = 0.47 mol ≈ 0.95 g left.
Problem 3 (Easy): Fe + 3Cl₂ → 2FeCl₃
11.2 g Fe and 21.3 g Cl₂. Which is limiting?
Cl₂: 21.3 g / 71.0 ≈ 0.300 mol ÷ 3 = 0.100
Cl₂ gives smaller number → Cl₂ is limiting.
Problem 4 (Medium): 2Na + Cl₂ → 2NaCl
34.5 g Na reacts with 42.6 g Cl₂. How many grams of NaCl are produced?
Cl₂: 42.6 / 71.0 ≈ 0.600 mol ÷ 1 = 0.600
Cl₂ is limiting.
Theoretical yield NaCl = 0.600 mol Cl₂ × (2 mol NaCl / 1 mol Cl₂) × 58.44 g/mol = 70.1 g
Problem 5 (Medium): C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
22.0 g propane (C₃H₈) and 80.0 g oxygen. How much excess oxygen remains?
O₂ needed = 0.500 × 5 = 2.50 mol
O₂ given = 80.0 / 32.0 = 2.50 mol → exactly stoichiometric!
No excess oxygen; both are completely consumed.
Problem 6 (Medium): 3Mg + N₂ → Mg₃N₂
36.5 g Mg and 14.0 g N₂. How many grams of excess reagent remain?
N₂: 14.0 / 28.0 = 0.500 mol ÷ 1 = 0.500
Both give ~0.50 → N₂ is slightly limiting (0.500 vs 0.501).
Mg used = 0.500 × 3 = 1.50 mol → 1.50 × 24.3 = 36.45 g used
Excess Mg = 36.5 – 36.45 = 0.05 g
Problem 7 (Hard): Al₂(SO₄)₃ + 3BaCl₂ → 3BaSO₄ + 2AlCl₃
85.5 g Al₂(SO₄)₃ and 104.2 g BaCl₂ are mixed.
(a) Limiting reagent? (b) Mass of BaSO₄ formed?
Al₂(SO₄)₃: 85.5 / 342 ≈ 0.250 mol ÷ 1 = 0.250
BaCl₂: 104.2 / 208.2 ≈ 0.500 mol ÷ 3 ≈ 0.167
BaCl₂ is limiting.
BaSO₄ produced = 0.500 mol BaCl₂ × (3 mol BaSO₄ / 3 mol BaCl₂) × 233.4 = 116.7 g
Problem 8 (Hard): 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
66.2 g of lead(II) nitrate decomposes. What mass of O₂ is produced?
Moles Pb(NO₃)₂ = 66.2 / 331.2 ≈ 0.200 mol
O₂ produced = 0.200 × (1 mol O₂ / 2 mol Pb(NO₃)₂) × 32.0 = 3.20 g O₂
Problem 9 (Challenge): 4NH₃ + 5O₂ → 4NO + 6H₂O
51.0 g NH₃ and 96.0 g O₂. Find:
(a) Limiting reagent
(b) Grams of NO produced
(c) Grams of excess reagent left
O₂: 96.0 / 32.0 = 3.00 mol ÷ 5 = 0.600
O₂ is limiting.
(a) Limiting = O₂
(b) NO produced = 3.00 mol O₂ × (4 mol NO / 5 mol O₂) × 30.0 = 72.0 g NO
(c) NH₃ used = 3.00 mol O₂ × (4 mol NH₃ / 5 mol O₂) = 2.40 mol → 2.40 × 17.0 = 40.8 g used
Excess NH₃ = 51.0 – 40.8 = 10.2 g left
Home Work
25.0 g Mg reacts with 20.0 g O₂ → 2MgO. How many grams of excess reagent remain?
Mg: 25.0 / 24.3 ≈ 1.03 mol ÷ 2 = 0.515
O₂: 20.0 / 32.0 = 0.625 mol ÷ 1 = 0.625 → Mg limiting
O₂ consumed = 1.03 × (1/2) = 0.515 mol
O₂ left = 0.625 – 0.515 = 0.11 mol → 3.5 g O₂ excess
Keep practicing! The more problems you solve, the faster you'll master Limiting Reagents, Percent Yield, and Excess Reagents.