This free limiting reactant worksheet walks Grade 11 students through stoichiometry one step at a time: mole ratios, mass-to-mass problems, limiting and excess reactants, and percent yield.
Each problem prints its balanced equation, and the answer key shows the full grams → moles → ratio → grams chain. Every equation has been checked atom by atom, and every answer has been calculated and verified to 3 significant figures.
- Grade 11 Chemistry
- 27 marks
- 45–60 minutes
- Worked answer key
What's in This Limiting Reactant Worksheet
The sheet builds the full stoichiometry toolkit before the limiting reactant problems, so students aren't learning everything at once.
| Part | Skill practised | Marks |
|---|
| A | Mole ratios from balanced equations | 4 |
| B | Mass-to-mass stoichiometry | 8 |
| C | Limiting reactant, product mass and excess reactant | 7 |
| D | Percent yield | 4 |
| E | Spot the mistake | 2 |
| F | Challenge: yields over 100% and propane combustion | 2 |
It supports the Ontario Grade 11 chemistry curriculum expectations in the Quantities in Chemical Reactions unit.
The Four Steps of Stoichiometry
Every calculation on this limiting reactant worksheet follows the same path:
grams of A → ÷ molar mass → moles of A → × mole ratio → moles of B → × molar mass → grams of B
The mole ratio comes from the coefficients in the balanced equation, which is why students need to be confident with the balancing chemical equations worksheet and the mole conversion worksheet first.
How to Find the Limiting Reactant
The limiting reactant runs out first and decides how much product can form. Here is question 9 worked through:
- Moles of each reactant: 4.00 g H2 ÷ 2.02 = 1.98 mol; 16.0 g O2 ÷ 32.00 = 0.500 mol.
- Product each could make (2H2 + O2 → 2H2O): H2 could make 1.98 mol of water; O2 could make 1.00 mol.
- The smaller amount wins. O2 is limiting, so the theoretical yield is 18.0 g of water.
Notice that hydrogen has the smaller mass but is not limiting. Comparing grams instead of moles is the most common mistake on any limiting reactant worksheet. For more background, see limiting reagent on Wikipedia.
Percent Yield
percent yield = actual yield ÷ theoretical yield × 100
Part D of the limiting reactant worksheet practises this. The theoretical yield is what the limiting reactant could make in a perfect reaction. Real yields are lower because of incomplete reactions, side reactions and losses during transfer.
Limiting Reactant and Stoichiometry Practice Problems
These match the limiting reactant worksheet PDF. Give answers to 3 significant figures, using the molar masses printed on the sheet.
Part A – Mole Ratios
- How many moles of O2 are needed to react with 4.0 mol of H2?
2H2 + O2 → 2H2O
- How many moles of H2O form from 3.0 mol of O2?
2H2 + O2 → 2H2O
- How many moles of NH3 form from 2.5 mol of N2?
N2 + 3H2 → 2NH3
- How many moles of H2 are needed to make 8.0 mol of NH3?
N2 + 3H2 → 2NH3
Part B – Mass-to-Mass Stoichiometry
- What mass of MgO forms from 12.0 g of Mg?
2Mg + O2 → 2MgO
- What mass of NH3 forms from 14.0 g of N2?
N2 + 3H2 → 2NH3
- What mass of CO2 forms from 8.00 g of CH4?
CH4 + 2O2 → CO2 + 2H2O
- What mass of O2 forms when 24.5 g of KClO3 decomposes?
2KClO3 → 2KCl + 3O2
Part C – Limiting Reactant
- 4.00 g of H2 reacts with 16.0 g of O2. Which is the limiting reactant, and what mass of H2O forms?
2H2 + O2 → 2H2O
- 28.0 g of N2 reacts with 9.00 g of H2. Which is the limiting reactant, and what mass of NH3 forms? How many grams of H2 are left over?
N2 + 3H2 → 2NH3
- 10.0 g of Al reacts with 35.0 g of Cl2. Which is the limiting reactant, and what mass of AlCl3 forms?
2Al + 3Cl2 → 2AlCl3
Part D – Percent Yield
- The theoretical yield of a reaction is 18.0 g, and 15.3 g is collected. What is the percent yield?
- In question 5, the theoretical yield of MgO is 19.9 g. A student collects 17.5 g. What is the percent yield?
- Heating 50.0 g of CaCO3 produces 25.2 g of CaO (CaCO3 → CaO + CO2). Find the percent yield.
Part E – Spot the Mistake
- For 2H2 + O2 → 2H2O, a student says 4.0 g of H2 reacts with 2.0 g of O2 “because the ratio is 2 : 1.” Explain the error and correct it.
- A student decides the reactant with the smaller mass must be the limiting reactant. Explain the error and correct it.
Part F – Challenge
- A student reports a percent yield of 108%. Give two likely reasons.
- Propane burns: C3H8 + 5O2 → 3CO2 + 4H2O. What mass of CO2 forms from 44.0 g of propane?
Limiting Reactant Worksheet Answer Key
Each answer shows the full chain of steps. The printable key is on pages 3–4 of the PDF.
Part A answers (Q1–4)
- 4.0 × 1/2 = 2.0 mol
- 3.0 × 2/1 = 6.0 mol
- 2.5 × 2/1 = 5.0 mol
- 8.0 × 3/2 = 12 mol
Part B answers (Q5–8)
- 12.0 g ÷ 24.31 = 0.494 mol Mg; × 2/2 = 0.494 mol MgO; × 40.31 = 19.9 g MgO
- 14.0 g ÷ 28.02 = 0.500 mol N2; × 2/1 = 0.999 mol NH3; × 17.04 = 17.0 g NH3
- 8.00 g ÷ 16.05 = 0.498 mol CH4; × 1/1 = 0.498 mol CO2; × 44.01 = 21.9 g CO2
- 24.5 g ÷ 122.55 = 0.200 mol KClO3; × 3/2 = 0.300 mol O2; × 32.00 = 9.60 g O2
Part C answers (Q9–11)
- H2: 1.98 mol → 1.98 mol H2O; O2: 0.500 mol → 1.00 mol H2O. Limiting: O2; mass of H2O = 18.0 g
- N2: 0.999 mol → 2.00 mol NH3; H2: 4.46 mol → 2.97 mol NH3. Limiting: N2; mass of NH3 = 34.1 g; H2 left over = 2.94 g
- Al: 0.371 mol → 0.371 mol AlCl3; Cl2: 0.494 mol → 0.329 mol AlCl3. Limiting: Cl2; mass of AlCl3 = 43.9 g
Part D answers (Q12–14)
- 15.3 ÷ 18.0 × 100 = 85.0%
- 17.5 ÷ 19.9 × 100 = 87.9%
- theoretical = 50.0 ÷ 100.09 × 56.08 = 28.0 g; 25.2 ÷ 28.0 × 100 = 90.0%
Part E answers (Q15–16)
- Coefficients are mole ratios, not mass ratios. 4.0 g H2 = 1.98 mol, which needs 0.990 mol O2 = 31.7 g O2.
- Compare moles using the mole ratio, not grams. In question 9, H2 has the smaller mass (4.00 g vs 16.0 g), but O2 is limiting.
Part F answers (Q17–18)
- The product was still wet (water adds mass) or contained impurities or unreacted solid; a weighing error is also possible.
- 44.0 g ÷ 44.11 = 0.998 mol C3H8; × 3/1 = 2.99 mol CO2; × 44.01 = 132 g CO2
5 Common Stoichiometry Mistakes
- Using coefficients as mass ratios. 2H2 + O2 means 2 mol to 1 mol, not 2 g to 1 g.
- Picking the limiting reactant by mass. Always compare moles of product each reactant can make.
- Flipping the mole ratio. Put the substance you want on top: mol B = mol A × (coefficient of B ÷ coefficient of A).
- Using the excess reactant for theoretical yield. Only the limiting reactant sets the maximum product.
- Forgetting diatomic molar masses. O2 is 32.00 g/mol, not 16.00.
How to Use This Limiting Reactant Worksheet in Class
- Day 1: Parts A and B to practise the four-step method.
- Day 2: Part C, working question 9 together before students try 10 and 11.
- Day 3: Parts D–F before a percent-yield lab, so students know what their numbers mean. Students who finish the whole limiting reactant worksheet are ready for solution stoichiometry.
For more practice, try the stoichiometry quiz and the stoichiometry calculation quiz, both auto-graded.
Frequently Asked Questions
Is this limiting reactant worksheet free?
Yes. The worksheet and its worked answer key are free to download and print as a PDF, with no account needed.
How do you find the limiting reactant?
Convert each reactant's mass to moles, then use the mole ratio to work out how much product each one could make. The reactant that makes less product is the limiting reactant.
Is the limiting reactant always the one with the smaller mass?
No. Compare moles using the balanced equation, not grams. In question 9, hydrogen has the smaller mass but oxygen is limiting.
How do you calculate percent yield?
Divide the actual yield by the theoretical yield and multiply by 100. The theoretical yield comes from the limiting reactant.
Can percent yield be over 100%?
Not for a pure, dry product. A result over 100% usually means the product was still wet or contained impurities.
More Free Grade 11 Chemistry Resources
Ready to practise? Download the free limiting reactant worksheet PDF (use the Download button above), print it double-sided, and keep the answer key pages for marking.