How to Do Stoichiometry — Limiting Reagent & Yields
Introduction
Stoichiometry is the quantitative heartbeat of chemistry—the discipline that transforms symbolic chemical equations into real-world predictions about how much reactant you need and how much product you’ll obtain. Whether you’re synthesising a life-saving drug in a lab, optimising an industrial chemical process, or simply completing a GCSE or A-Level assignment, mastering stoichiometry is non-negotiable.
At its core, stoichiometry relies on the law of conservation of mass and the mole concept to convert between masses, moles, and molecules using the ratios provided by a balanced chemical equation. Yet many students struggle—not because the maths is complex, but because they miss the logical sequence that connects these steps. This guide demystifies the process, walking you through how to balance equations, identify the limiting reagent, calculate theoretical and percent yields, and avoid common pitfalls that lead to incorrect answers. You’ll learn not just how to do stoichiometry, but why each step matters, turning a source of anxiety into a powerful problem-solving skill.
The Stoichiometric Workflow: From Equation to Answer
Every stoichiometry problem follows a consistent, logical pathway. Master this sequence, and you can solve any problem—from simple mass-to-mass conversions to complex multi-reactant scenarios.
Step 1: Balance the Chemical Equation
This is your foundation. An unbalanced equation violates the law of conservation of mass and will give you entirely wrong mole ratios.
Example:
Unbalanced: Fe + O₂ → Fe₂O₃
Balanced: 4Fe + 3O₂ → 2Fe₂O₃
Now the mole ratio of Fe to O₂ to Fe₂O₃ is 4 : 3 : 2—this is your conversion roadmap.
Step 2: Convert Given Quantities to Moles
Never work directly with grams in stoichiometric ratios. Moles are the universal currency of chemical reactions.
Use:
- Molar mass (g/mol) for solids/liquids:
moles = mass (g) / molar mass (g/mol) - Molarity and volume for solutions:
moles = concentration (mol/L) × volume (L) - Molar volume for gases at STP:
moles = volume (L) / 22.4 L/mol(UK exams often use 24.0 L/mol at RTP—check your syllabus!)
Step 3: Apply the Mole Ratio
Use coefficients from the balanced equation as conversion factors.
From 4Fe + 3O₂ → 2Fe₂O₃:
- To find moles of Fe₂O₃ from Fe:
mol Fe × (2 mol Fe₂O₃ / 4 mol Fe) - To find moles of O₂ needed for a given Fe:
mol Fe × (3 mol O₂ / 4 mol Fe)
Step 4: Convert Moles to Desired Units
Finally, convert back to grams, litres, or molecules as required by the question.
Identifying the Limiting Reagent
When two or more reactants are given, one will run out first—the limiting reagent. It determines the maximum possible product.
Method:
- Convert all reactants to moles.
- Calculate how much product each could produce.
- The reactant that yields the least product is limiting.
Why it matters: Excess reagents are wasted in industrial processes—identifying the limiting reagent optimises cost and efficiency.
Calculating Yields
- Theoretical yield: Maximum product possible (from limiting reagent).
- Actual yield: Amount obtained in lab (always ≤ theoretical).
- Percent yield:
(Actual / Theoretical) × 100%
A high percent yield (>90%) indicates a clean, efficient reaction; low yields suggest side reactions, incomplete processes, or loss during transfer.
Handling Hydrates and Gases
- Hydrates (e.g., CuSO₄·5H₂O): Include water mass in molar mass.
- Gases: Specify conditions—STP (0°C, 1 atm) vs. RTP (25°C, 1 atm)—as molar volume differs (22.4 L/mol vs. 24.0 L/mol).
Worked Examples & Practice Problems
Example 1: Basic Mass-to-Mass Conversion
Problem: What mass of CO₂ is produced from burning 50.0 g of propane (C₃H₈)?
Balanced: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Solution:
- Molar mass C₃H₈ = 44.10 g/mol →
50.0 / 44.10 = 1.134 mol - Mole ratio:
1.134 mol C₃H₈ × (3 mol CO₂ / 1 mol C₃H₈) = 3.402 mol CO₂ - Molar mass CO₂ = 44.01 g/mol →
3.402 × 44.01 = 149.7 g
✅ Answer: 150 g CO₂ (3 sig figs)
Example 2: Limiting Reagent & Percent Yield
Problem: 15.0 g Al reacts with 25.0 g Cl₂ to form AlCl₃. Actual yield = 28.5 g. Find theoretical yield and % yield.
Balanced: 2Al + 3Cl₂ → 2AlCl₃
Solution:
- Moles Al =
15.0 / 26.98 = 0.556 mol
Moles Cl₂ =25.0 / 70.90 = 0.353 mol - Product from Al:
0.556 × (2/2) = 0.556 mol AlCl₃
Product from Cl₂:0.353 × (2/3) = 0.235 mol AlCl₃→ Cl₂ is limiting - Theoretical yield =
0.235 mol × 133.33 g/mol = 31.3 g - % yield =
(28.5 / 31.3) × 100% = 91.1%
Example 3: Solution Stoichiometry
Problem: What volume of 0.500 M NaOH is needed to neutralise 25.0 mL of 0.200 M H₂SO₄?
Balanced: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Solution:
- Moles H₂SO₄ =
0.0250 L × 0.200 mol/L = 0.00500 mol - Mole ratio:
0.00500 mol H₂SO₄ × (2 mol NaOH / 1 mol H₂SO₄) = 0.0100 mol NaOH - Volume NaOH =
0.0100 mol / 0.500 mol/L = 0.0200 L = 20.0 mL
Practice Problems (Try These!)
- How many grams of O₂ are needed to burn 32.0 g of CH₄?
- If 10.0 g Mg and 10.0 g O₂ react, what mass of MgO forms?
- What is the % yield if 8.50 g of NH₃ produces 12.0 g of NO in the Ostwald process?
Answers:
- 128 g
- 16.6 g (Mg limiting)
- 80.0%
Why must the equation be balanced first?
Because stoichiometric ratios come directly from the coefficients. An unbalanced equation gives false ratios—like using a map with wrong distances.
Can I skip converting to moles?
No. Grams don’t have a direct ratio in reactions—only moles do. Skipping this step is the #1 error in stoichiometry.
How do I know which reactant is limiting without calculating both?
You can compare the mole ratio available vs. mole ratio required.
For 2A + 3B → C, if you have 4 mol A and 5 mol B:
- Required B/A = 3/2 = 1.5
- Available B/A = 5/4 = 1.25 below 1.5 → B is limiting
Why is my percent yield over 100%?
Usually due to impure product (e.g., still wet with solvent) or measurement error. True yields cannot exceed 100%.
Do I include water in hydrate molar mass?
Yes! For CuSO₄·5H₂O, molar mass = 63.55 + 32.07 + 64.00 + 5×(18.02) = 249.7 g/mol.
What if the problem gives volume of a gas not at STP?
Use the ideal gas law: PV = nRT to find moles first, then proceed.
How many significant figures should I use?
Base your final answer on the least precise measurement in the problem. Carry extra digits through intermediate steps.
Is stoichiometry used outside of exams?
Absolutely! It’s essential in pharmaceutical manufacturing, environmental engineering (e.g., calculating scrubber efficiency), food science, and materials production.