Find limiting reagents and stoichiometric relationships
Compare reactant amounts against a balanced equation, identify excess material, and deduce the reacting ratio from measured results.
Section 2 · Atoms, molecules and stoichiometry
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Use the data and constants supplied for the question, show mole-ratio working, and report a justified number of significant figures. Written practice is self-checked
Before you begin
- Use common chemical symbols and formulae, and rearrange a simple equation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which reactant determines the theoretical product amount?
What you will learn
- Identify limiting and excess reagents using amount ratios
- Use reaction data to deduce stoichiometric relationships
The limiting reagent is the one exhausted first
For aA + bB → products, compare n(A)/a with n(B)/b. The smaller ratio identifies the limiting reagent; the other reactant is in excess. The limiting reactant sets the maximum product amount.
Do not compare reactant masses or raw mole amounts without using the equation coefficients. If needed, calculate excess left over by subtracting the amount consumed from the initial amount. A negative leftover signals an arithmetic or limiting-reagent error.
Experimental amounts can reveal a reaction ratio
To deduce a stoichiometric relationship from measurements, convert each measured mass, gas volume or solution volume to amount using the relevant formula and stated conditions. Reduce the amount ratio to the simplest whole numbers, then check that the proposed coefficients fit the observed data.
Real measurements may be affected by uncertainty or incomplete reaction. State the inferred relationship from the supplied data and do not treat a rounded experimental ratio as exact unless the evidence supports that precision.
Pause and explain
For 2A + B → products, which comparison identifies the limiting reagent?
Worked example
N₂ + 3H₂ → 2NH₃. A mixture contains 2.0 mol N₂ and 5.0 mol H₂. Identify the limiting reagent and maximum amount of NH₃.
Show the worked solution
- For N₂, n/coefficient = 2.0/1 = 2.0.
- For H₂, n/coefficient = 5.0/3 = 1.67; this is smaller.
- H₂ is limiting and produces 2×(5.0/3) mol NH₃.
Answer H₂ is limiting; the maximum is 3.33 mol NH₃.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
2Al + 3Cl₂ → 2AlCl₃. A vessel contains 0.40 mol Al and 0.50 mol Cl₂. Identify the limiting reagent and maximum moles of AlCl₃.
Give me a hint
Divide each amount by its coefficient, then use the smaller value.
Compare my reasoning
- Al: 0.40/2 = 0.20.
- Cl₂: 0.50/3 ≈ 0.167, so Cl₂ is limiting.
- AlCl₃ amount = 2×0.167.
Cl₂ is limiting; maximum AlCl₃ ≈ 0.333 mol.
Look for these in your work
- Coefficient-normalised comparison
- Limiting reagent identified
- Product coefficient used
An experiment reacts 0.30 mol magnesium with 0.15 mol oxygen to form magnesium oxide. Deduce the simplest mole ratio Mg:O₂ and write the balanced equation.
Give me a hint
Simplify the measured amounts, then check that the product formula conserves oxygen atoms.
Compare my reasoning
- The measured ratio is 0.30:0.15.
- Divide both amounts by 0.15 to obtain 2:1.
- Use the ratio as coefficients and check the oxygen atoms.
Mg:O₂ = 2:1; the balanced equation is 2Mg + O₂ → 2MgO.
Look for these in your work
- Measured ratio reduced correctly
- Equation coefficients match measured ratio
- Atom balance checked
A learner says 0.20 mol H₂ is limiting in 2H₂ + O₂ → 2H₂O when 0.10 mol O₂ is present. Assess the claim and state how much water can form.
Give me a hint
Compare each amount divided by its coefficient.
Compare my reasoning
- H₂: 0.20/2 = 0.10.
- O₂: 0.10/1 = 0.10.
- The reactants are in the exact stoichiometric ratio.
Neither reagent is in excess: the amounts are exactly stoichiometric. The reaction can form 0.20 mol H₂O, and both reactants are exhausted together.
Look for these in your work
- Normalised amounts compared
- No false limiting reagent asserted
- Product amount follows coefficient ratio
Common mistakes
- Selecting the reactant with the smaller mass
- Ignoring coefficients when comparing available moles
- Calling the excess reagent the limiting reagent
- Treating an approximate experimental ratio as exact beyond its precision
What can you explain now?
Which reactant determines the theoretical product amount?
Compare with the explanation
The limiting reagent
It is exhausted first and sets the maximum product amount.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How does coefficient-normalised amount reveal the limiting reagent, and how should experimental ratios be checked?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.