The shape of the lesson
By the end of the lesson, learners can
- The three-step route: Never compare grams directly: 1 g of one substance and 1 g of another contain different numbers of particles.
- Limiting and excess: Set the two amounts in the simulator and watch which bar empties first.
- Percentage yield: Change the mass of reactant and the mass of product actually collected, and read the percentage yield.
- Mass → moles: n = m ÷ M.
- Ratio: from the balanced equation, never from masses.
- Limiting reagent: smallest (moles ÷ coefficient).
- % yield = actual ÷ theoretical (from the limiting reagent) × 100.
One picture every idea returns to
Mass of A → moles of A (divide by M) → moles of B (use the equation's ratio) → mass of B (multiply by M). The limiting reagent is the reactant that is completely used up; it sets the maximum amount of product, and the other reactant is in excess. Percentage yield = actual yield ÷ theoretical yield × 100, with the theoretical yield calculated from the limiting reagent.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The equation counts particles.. A balanced equation gives a ratio of moles, not of grams. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The three-step route | Never compare grams directly: 1 g of one substance and 1 g of another contain different numbers of particles. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Limiting and excess | Set the two amounts in the simulator and watch which bar empties first. Project tab 4 of the lab (Stoichiometry lab, limiting reagent) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Percentage yield | Change the mass of reactant and the mass of product actually collected, and read the percentage yield. Project tab 7 of the lab (Stoichiometry lab, yield) and let a student drive it while the class predicts. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 17 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Percentage yield, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Use the ratio from the equation, not a ratio of masses, and always work the theoretical yield from the limiting reagent. | If you use the reagent in excess, the answer will be too big and the percentage yield too small. |
Reacting masses and limiting reagents: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Calculate the mass of carbon dioxide formed when 5.00 g of calcium carbonate reacts with excess hydrochloric acid. CaCO3 + 2HCl → CaCl2 + H2O + CO2. (Mr: CaCO3 100.1, CO2 44.0)
5.40 g of aluminium is heated with 14.2 g of chlorine. 2Al + 3Cl2 → 2AlCl3. Identify the limiting reagent and calculate the mass of aluminium chloride formed and the mass of the reagent left over. (Ar: Al 27.0, Cl 35.5)
12.0 g of calcium carbonate is heated and 5.60 g of calcium oxide is collected. Calculate the percentage yield. (Mr: CaCO3 100.1, CaO 56.1)
1.12 g of iron reacts completely with 2.13 g of chlorine, Cl2, to form a single chloride. Use the mole ratio to deduce the balanced equation. (Ar: Cl 35.5, Fe 55.8)
An iron ore contains 80.0% Fe2O3 by mass. Calculate the maximum mass of iron, in tonnes, that can be extracted from 1.00 tonne of the ore. Fe2O3 + 3CO → 2Fe + 3CO2. (Mr: Fe2O3 159.6; Ar: Fe 55.8; 1 tonne = 106 g)
Suggest three reasons why the actual yield of a product in a laboratory preparation is less than the theoretical yield.
Answer key and marking notes
- n(CaCO3) = 5.00 ÷ 100.1 = 0.0500 mol, so n(CO2) = 0.0500 mol.
- Mass of CO2 = 0.0500 × 44.0 = 2.20 g.
- n(Al) = 5.40 ÷ 27.0 = 0.200 mol; n(Cl2) = 14.2 ÷ 71.0 = 0.200 mol.
- Dividing by the coefficients: Al 0.200 ÷ 2 = 0.100; Cl2 0.200 ÷ 3 = 0.0667. Chlorine is limiting.
- n(AlCl3) = 0.200 × 2/3 = 0.133 mol; Mr = 133.5, so mass = 17.8 g.
- Al used = 0.133 mol, so 0.0667 mol is left: 0.0667 × 27.0 = 1.80 g of aluminium in excess.
- Theoretical: n = 12.0 ÷ 100.1 = 0.120 mol, so 0.120 × 56.1 = 6.73 g of CaO.
- Percentage yield = 5.60 ÷ 6.73 × 100 = 83.3%.
- n(Fe) = 1.12 ÷ 55.8 = 0.0201 mol; n(Cl2) = 2.13 ÷ 71.0 = 0.0300 mol.
- Fe : Cl2 = 0.0201 : 0.0300 = 2 : 3.
- 2Fe + 3Cl2 → 2FeCl3.
- Mass of Fe2O3 = 0.800 × 106 g, so n = 0.800 × 106 ÷ 159.6 = 5.01 × 103 mol.
- n(Fe) = 2 × 5.01 × 103 = 1.00 × 104 mol.
- Mass = 1.00 × 104 × 55.8 = 5.59 × 105 g = 0.559 tonnes.
- The reaction may not go to completion, or may be reversible and reach equilibrium.
- Side reactions may convert some of the reactant into other products.
- Some product is lost during transfer, filtration, recrystallisation or other purification steps.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
6.08 g of magnesium is burned in excess oxygen: 2Mg + O2 → 2MgO. What mass of magnesium oxide forms? (Ar: O 16.0, Mg 24.3)
- A10.1 g
- B5.04 g
- C20.2 g
- D14.1 g
What mass of oxygen is needed to burn 8.0 g of methane completely? CH4 + 2O2 → CO2 + 2H2O (Ar: H 1.0, C 12.0, O 16.0)
- A32.0 g
- B16.0 g
- C64.0 g
- D1.00 g
4.60 g of sodium reacts with water: 2Na + 2H2O → 2NaOH + H2. After evaporation, 6.40 g of sodium hydroxide is obtained. What is the percentage yield? (Ar: H 1.0, O 16.0, Na 23.0)
- A80.0%
- B139%
- C71.9%
- D160%
2.0 mol of nitrogen is mixed with 3.0 mol of hydrogen and they react completely: N2 + 3H2 → 2NH3. What is the maximum amount of ammonia that can form?
- A2.0 mol
- B4.0 mol
- C1.3 mol
- D5.0 mol
Excess iron powder is added to a solution containing 0.0500 mol of copper(II) sulfate. Fe + CuSO4 → FeSO4 + Cu. What mass of copper is displaced? (Ar: O 16.0, S 32.1, Fe 55.8, Cu 63.5)
- A3.18 g
- B7.98 g
- C2.79 g
- D6.35 g
10.0 g of iron is heated with 10.0 g of sulfur: Fe + S → FeS. 12.6 g of iron(II) sulfide (Mr 87.9) is obtained. What is the percentage yield? (Ar: S 32.1, Fe 55.8)
- A80.0%
- B46.0%
- C63.0%
- D126%
Ethyl ethanoate (Mr 88.0) is made from ethanol (Mr 46.0): CH3COOH + C2H5OH → CH3COOC2H5 + H2O. The percentage yield is 60.0%. What mass of ethanol must be used to obtain 5.00 g of ethyl ethanoate?
- A4.36 g
- B1.57 g
- C2.61 g
- D8.33 g
Different volumes of 0.100 mol dm−3 barium chloride are added to separate 25.0 cm3 portions of sodium sulfate solution. Each precipitate of barium sulfate (Mr 233.4) is filtered, dried and weighed. The results are shown. Ba2+(aq) + SO42−(aq) → BaSO4(s). What is the concentration of the sodium sulfate solution?
- A0.0800 mol dm−3
- B0.100 mol dm−3
- C0.125 mol dm−3
- D8.00 × 10−5 mol dm−3
Quiz key
n(Mg) = 6.08 ÷ 24.3 = 0.250 mol. Mg : MgO = 2 : 2 = 1 : 1, so 0.250 mol MgO, and mass = 0.250 × 40.3 = 10.1 g. (Check: 0.125 mol O2 = 4.00 g, and 6.08 + 4.00 = 10.1 g.) 5.04 g halves the amount, reading the 2 in front of MgO as a 1 : 2 ratio. 20.2 g doubles it. 14.1 g adds 8.00 g of oxygen, assuming one O2 per Mg instead of one per two Mg.
n(CH4) = 8.0 ÷ 16.0 = 0.50 mol. The ratio is 1 : 2, so n(O2) = 1.00 mol and mass = 1.00 × 32.0 = 32.0 g. 16.0 g ignores the 2 in front of O2. 64.0 g treats the 8.0 g of methane as a whole mole. 1.00 g is the amount in moles written as if it were a mass.
n(Na) = 4.60 ÷ 23.0 = 0.200 mol, and Na : NaOH = 1 : 1, so the theoretical yield is 0.200 × 40.0 = 8.00 g. Percentage yield = 6.40 ÷ 8.00 × 100 = 80.0%. 139% divides the mass of product by the mass of sodium, a mass ratio with no chemical meaning. 71.9% is the same mass ratio upside down. 160% takes 0.100 mol NaOH, using the Na : H2 ratio by mistake.
2.0 mol N2 would need 6.0 mol H2, but only 3.0 mol is present, so hydrogen is limiting. H2 : NH3 = 3 : 2, giving 3.0 × 2/3 = 2.0 mol NH3 (with 1.0 mol N2 left over). 4.0 mol assumes nitrogen is limiting. 1.3 mol applies the 3 : 2 ratio to the 2.0 mol of nitrogen. 5.0 mol simply adds the reactant amounts.
CuSO4 : Cu = 1 : 1, so 0.0500 mol Cu forms, with mass 0.0500 × 63.5 = 3.18 g. 7.98 g is the mass of the copper(II) sulfate (0.0500 × 159.6), not of the copper. 2.79 g is the mass of iron used. 6.35 g doubles the amount of copper although the ratio is 1 : 1.
n(Fe) = 10.0 ÷ 55.8 = 0.179 mol; n(S) = 10.0 ÷ 32.1 = 0.312 mol. The ratio is 1 : 1, so iron is limiting. Theoretical FeS = 0.179 × 87.9 = 15.8 g, and yield = 12.6 ÷ 15.8 × 100 = 80.0%. 46.0% bases the theoretical yield on sulfur (27.4 g), which is in excess. 63.0% compares the product with the total mass of reactants (20.0 g), ignoring the unreacted sulfur. 126% divides by the mass of iron alone.
n(ester) = 5.00 ÷ 88.0 = 0.0568 mol. At 60.0% yield the ethanol needed is 0.0568 ÷ 0.600 = 0.0947 mol, so mass = 0.0947 × 46.0 = 4.36 g. 1.57 g multiplies by 0.600 instead of dividing, which would give even less ester. 2.61 g ignores the yield (the 100% answer). 8.33 g divides the mass of ester by 0.600 and skips the moles, ignoring the different molar masses.
The line levels off at 20.0 cm3: beyond that, sulfate is the limiting reagent. n(Ba2+) = 0.100 × 20.0/1000 = 0.00200 mol = n(SO42−) (check: 0.467 ÷ 233.4 = 0.00200 mol). Concentration = 0.00200 ÷ 0.0250 = 0.0800 mol dm−3. 0.100 mol dm−3 is the barium chloride concentration, not the sodium sulfate. 0.125 mol dm−3 puts the volumes the wrong way round (0.100 × 25.0 ÷ 20.0). 8.00 × 10−5 mol dm−3 divides by 25.0 cm3 without converting it to dm3.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.