Stoichiometric ratios and limiting reactants
Let the balanced equation connect amounts, then identify which reactant runs out first.
Section 3.3 · The mole and the Avogadro constant
Exam scope and assessment notes
Extended only (Supplement) · Chemistry 0971, 2026–2028. Core reacting-mass proportions are taught separately. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Balance symbol equations and calculate molar masses.
- Convert between mass and amount using n=m/M.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does the coefficient three mean for H₂ in N₂+3H₂→2NH₃?
What you will learn
- Use balanced coefficients to connect reacting amounts and product masses.
- Identify the limiting reactant and calculate the excess remaining.
Coefficients connect moles, not grams
In N₂+3H₂→2NH₃, one mole of nitrogen molecules reacts with three moles of hydrogen molecules and forms two moles of ammonia molecules. The amount ratio is 1:3:2. To use masses, convert each mass to moles first, apply the coefficient ratio and convert the requested amount back to grams. Assume complete reaction for a theoretical prediction unless the question states otherwise.
Write the ratio in the direction you need. For 0.20 mol N₂ with sufficient hydrogen, ammonia amount is 0.20×2/1=0.40 mol. Multiplying grams directly by the coefficient ratio is generally wrong because the substances have different molar masses. Check the balanced equation before any numerical work.
Compare available reaction portions
When both reactants are given, divide each available amount in moles by its coefficient. The smaller result gives the number of complete reaction portions available and identifies the limiting reactant. Multiply that value by each product coefficient to obtain the theoretical product amount. If the values are equal, the supplied amounts are in the exact required ratio.
Calculate excess left by subtracting the amount actually needed from the amount supplied. For 1.0 mol N₂ and 2.0 mol H₂, the available portions are 1.0/1 and 2.0/3. Hydrogen limits, even though its numerical amount in moles is larger. The nitrogen consumed is 2.0/3 mol, leaving 1.0−2.0/3 mol. A limiting reactant is identified from both the amount and its coefficient.
Pause and explain
How do you compare reactants in an equation with unequal coefficients?
Worked example
8.0 g H₂ reacts with 28 g N₂ in N₂+3H₂→2NH₃. Find the limiting reactant, theoretical ammonia mass and excess H₂ left. Use H=1 and N=14.
Show the worked solution
- Amounts are H₂:8/2=4.0 mol and N₂:28/28=1.0 mol.
- Available portions are 4/3 and 1/1, so nitrogen limits; 1.0 mol N₂ uses 3.0 mol H₂ and forms 2.0 mol NH₃.
- Ammonia mass is 2×17=34 g. Hydrogen remaining is (4−3)×2=2 g. Reacted masses 28+6=34 g.
Answer Nitrogen limits; 34 g NH₃; 2 g H₂ remains.
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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.
For 2Mg+O₂→2MgO, 4.8 g Mg reacts with 6.4 g O₂. Find the limiting reactant and MgO mass, using Mg=24 and O=16.
Give me a hint
Convert both masses to moles and compare n/coefficient.
Compare my reasoning
- n(Mg)=4.8/24=0.20 mol; n(O₂)=6.4/32=0.20 mol.
- Portions are 0.20/2=0.10 and 0.20/1=0.20, so magnesium limits.
- MgO amount is 2×0.10=0.20 mol; mass is 0.20×40=8.0 g.
Magnesium limits; 8.0 g MgO.
Look for these in your work
- Both reactants are checked with coefficients.
- Product moles are converted using MgO's own molar mass.
For CaCO₃→CaO+CO₂, calculate both product amounts and masses from 10 g CaCO₃ assuming complete decomposition. Use Ca=40, C=12, O=16.
Give me a hint
The coefficients are all one, but the three molar masses differ.
Compare my reasoning
- CaCO₃ molar mass is 100 g/mol, so its amount is 10/100=0.10 mol.
- Both products have amount 0.10 mol.
- CaO mass is 0.10×56=5.6 g and CO₂ mass is 0.10×44=4.4 g, totalling 10 g.
0.10 mol of each product; 5.6 g CaO and 4.4 g CO₂.
Look for these in your work
- Equal amounts are distinguished from equal masses.
- All product masses account for the reactant mass.
For N₂+3H₂→2NH₃, 0.50 mol N₂ and 0.90 mol H₂ are supplied. Find ammonia amount and the nitrogen amount left.
Give me a hint
Hydrogen supplies 0.90/3 reaction portions.
Compare my reasoning
- Compare 0.50/1=0.50 with 0.90/3=0.30; hydrogen limits.
- Ammonia amount is 2×0.30=0.60 mol; nitrogen used is 1×0.30=0.30 mol.
- Nitrogen left is 0.50−0.30=0.20 mol.
0.60 mol NH₃; 0.20 mol N₂ remains.
Look for these in your work
- The prediction follows the limiting reactant.
- Excess is found by subtracting amounts of the same substance.
Common mistakes
- Choosing the reactant with the smaller mass without considering molar masses and coefficients.
- Predicting product from the excess reactant as though all of it reacted.
- Subtracting masses of different substances to find excess amount.
What can you explain now?
For 2H₂+O₂→2H₂O, 1.0 mol H₂ and 1.0 mol O₂ are supplied. Which limits and how much water can form?
Compare with the explanation
H₂ limits; 1.0 mol H₂O.
Compare 1.0/2 with 1.0/1. The smaller hydrogen value is 0.50 reaction portions, giving 2×0.50=1.0 mol water.
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Make it stick
Explain why the smaller number of moles does not always identify the limiting reactant. Use an equation with unequal coefficients and calculate both product and excess.
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