The shape of the lesson
By the end of the lesson, learners can
- Three steps through the mole: Extended content.
- The limiting reactant: Set the amounts of the two reactants in the simulator and watch which one runs out first.
- Yield and purity: Try the yield tab: real reactions lose product on the way, and real starting materials contain impurities.
- Step 1: moles = mass ÷ Mr (or gas volume ÷ 24).
- Step 2: use the numbers in the balanced equation; Mr never includes them.
- Step 3: mass = moles × Mr (or volume = moles × 24).
- Limiting reactant first, then calculate the product from it.
One picture every idea returns to
Use three steps: (1) change the mass (or gas volume) you are given into moles; (2) use the numbers in the balanced equation to find the moles of the substance you want; (3) change those moles back into a mass or volume. The limiting reactant is the one that is completely used up; it fixes the amount of product. Percentage yield = actual mass ÷ expected mass × 100. Percentage purity = mass of pure substance ÷ mass of impure sample × 100.
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 moles.. 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: Three steps through the mole | Extended content. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The limiting reactant | Set the amounts of the two reactants in the simulator and watch which one runs out 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: Yield and purity | Try the yield tab: real reactions lose product on the way, and real starting materials contain impurities. Project tab 7 of the lab (Stoichiometry lab, yield) and let a student drive it while the class predicts. | Rally Coach |
| 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 Yield and purity, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| The 2 in front of 2MgO belongs in the mole ratio, not in the M r. | Use M r (MgO) = 40, not 80. A student who calculates 0.2 mol × 80 gets 16 g, double the correct 8.0 g. |
| Equal numbers of moles do not mean nothing is left over. | Here there are 0.1 mol of each, but the equation needs twice as much HCl as Mg, so HCl runs out first. Always compare the moles you have with the ratio in the equation. |
Reacting masses, yield and purity: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Iron is made in the blast furnace: Fe2O3 + 3CO → 2Fe + 3CO2. Calculate the mass of iron that can be made from 320 g of iron(III) oxide. (Ar: O 16, Fe 56)
Calculate the volume of carbon dioxide, at r.t.p., given off when 25 g of calcium carbonate is heated until it has all decomposed. CaCO3 → CaO + CO2 (Ar: C 12, O 16, Ca 40)
6.5 g of zinc is added to 3.65 g of hydrogen chloride dissolved in water. Zn + 2HCl → ZnCl2 + H2. (a) Show which reactant is limiting. (b) Calculate the volume of hydrogen formed at r.t.p. (c) Calculate the mass of the reactant left over. (Ar: H 1, Cl 35.5, Zn 65)
A student burns 4.8 g of magnesium and expects to make 8.0 g of magnesium oxide. She collects 7.2 g. Calculate the percentage yield.
10.0 g of impure calcium carbonate reacts with excess hydrochloric acid and gives 2.16 dm3 of carbon dioxide at r.t.p. CaCO3 + 2HCl → CaCl2 + H2O + CO2. Calculate the percentage purity of the calcium carbonate. (Ar: C 12, O 16, Ca 40)
Give two reasons why the actual yield of a product is usually less than the expected yield.
Answer key and marking notes
- Mr(Fe2O3) = 160, so moles = 320 ÷ 160 = 2 mol.
- Ratio Fe2O3 : Fe = 1 : 2, so 4 mol of Fe.
- Mass of Fe = 4 × 56 = 224 g.
- Mr(CaCO3) = 100, so moles = 25 ÷ 100 = 0.25 mol.
- Ratio 1 : 1, so 0.25 mol of CO2.
- Volume = 0.25 × 24 = 6 dm3.
- Zn: 6.5 ÷ 65 = 0.1 mol. HCl (Mr 36.5): 3.65 ÷ 36.5 = 0.1 mol.
- 0.1 mol of Zn needs 0.2 mol of HCl, but only 0.1 mol is present, so HCl is limiting.
- 0.1 mol HCl gives 0.05 mol H2 = 0.05 × 24 = 1.2 dm3.
- 0.05 mol of Zn reacts, so 0.05 mol is left: 0.05 × 65 = 3.25 g of zinc.
- Percentage yield = actual ÷ expected × 100.
- 7.2 ÷ 8.0 × 100 = 90%.
- Moles of CO2 = 2.16 ÷ 24 = 0.09 mol.
- Ratio 1 : 1, so 0.09 mol of CaCO3 = 0.09 × 100 = 9.0 g.
- Purity = 9.0 ÷ 10.0 × 100 = 90%.
- The reaction may not go to completion (for example, it is reversible) or other side reactions make different products.
- Some product is lost when it is separated, filtered, transferred or dried.
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.
Sodium burns in chlorine: 2Na + Cl2 → 2NaCl. What mass of sodium chloride forms from 4.6 g of sodium? (Ar: Na 23, Cl 35.5)
- A11.7 g
- B23.4 g
- C5.85 g
- D7.1 g
Zinc reacts with excess dilute sulfuric acid: Zn + H2SO4 → ZnSO4 + H2. What volume of hydrogen forms at r.t.p. from 1.3 g of zinc? (Ar: Zn 65)
- A480 cm3
- B960 cm3
- C240 cm3
- D0.48 cm3
Copper(II) sulfate is made from copper(II) oxide: CuO + H2SO4 → CuSO4 + H2O. What mass of copper(II) oxide is needed to make 8.0 g of CuSO4? (Ar: O 16, S 32, Cu 64)
- A4.0 g
- B8.0 g
- C16 g
- D3.2 g
4 g of hydrogen is mixed with 16 g of oxygen and exploded: 2H2 + O2 → 2H2O. Which reactant is limiting, and what mass of water forms? (Ar: H 1, O 16)
- Aoxygen; 18 g
- Bhydrogen; 36 g
- Coxygen; 9 g
- Dhydrogen; 20 g
14 g of nitrogen reacts with excess hydrogen: N2 + 3H2 ⇌ 2NH3. Only 5.1 g of ammonia is collected. What is the percentage yield? (Ar: H 1, N 14)
- A30%
- B60%
- C36.4%
- D15%
2.5 g of impure magnesium reacts with excess hydrochloric acid, giving 1.8 dm3 of hydrogen at r.t.p. Mg + 2HCl → MgCl2 + H2. The impurities do not react. What is the percentage purity of the magnesium? (Ar: Mg 24)
- A72%
- B28%
- C36%
- D144%
Marble chips (calcium carbonate) react with excess hydrochloric acid and the carbon dioxide is collected at r.t.p. The graph shows the results. CaCO3 + 2HCl → CaCl2 + H2O + CO2. What mass of calcium carbonate reacted? (Ar: C 12, O 16, Ca 40)
- A1.0 g
- B0.44 g
- C10 g
- D0.10 g
Carbon monoxide burns: 2CO(g) + O2(g) → 2CO2(g). What volume of oxygen is needed to burn 60 cm3 of carbon monoxide completely? (All volumes at the same temperature and pressure.)
- A30 cm3
- B60 cm3
- C120 cm3
- D15 cm3
Quiz key
Na: 4.6 ÷ 23 = 0.2 mol. The ratio Na : NaCl is 2 : 2, so 0.2 mol of NaCl forms: 0.2 × 58.5 = 11.7 g. 23.4 g uses 117 (2 × 58.5) as the Mr; 5.85 g halves the moles; 7.1 g is the mass of chlorine used (0.1 × 71).
Zn: 1.3 ÷ 65 = 0.02 mol. Ratio 1 : 1, so 0.02 mol of H2. Volume = 0.02 × 24 000 = 480 cm3 (0.48 dm3). 960 cm3 doubles the moles because H2 has two atoms; 240 cm3 halves them; 0.48 is the value in dm3, not cm3.
Mr(CuSO4) = 64 + 32 + 64 = 160, so 8.0 ÷ 160 = 0.05 mol. Ratio 1 : 1, so 0.05 mol of CuO (Mr 80) = 0.05 × 80 = 4.0 g. 8.0 g assumes the masses react 1 : 1; 16 g multiplies by 160 ÷ 80 instead of 80 ÷ 160; 3.2 g uses the Mr of copper alone.
H2: 4 ÷ 2 = 2 mol. O2: 16 ÷ 32 = 0.5 mol. 2 mol of H2 would need 1 mol of O2, but only 0.5 mol is present, so oxygen is limiting. 0.5 mol O2 gives 2 × 0.5 = 1 mol H2O = 18 g. 36 g treats hydrogen as limiting; 9 g forgets the 1 : 2 ratio; 20 g just adds the masses, as if both reactants were used up.
N2: 14 ÷ 28 = 0.5 mol, giving 2 × 0.5 = 1 mol NH3 = 17 g expected. Yield = 5.1 ÷ 17 × 100 = 30%. 60% uses a 1 : 1 ratio (8.5 g expected); 36.4% divides by the mass of nitrogen; 15% uses 34 as the Mr of 2NH3.
H2: 1.8 ÷ 24 = 0.075 mol. Mg : H2 is 1 : 1, so 0.075 mol Mg = 0.075 × 24 = 1.8 g. Purity = 1.8 ÷ 2.5 × 100 = 72%. 28% is the percentage of impurity; 36% halves the moles of Mg; 144% doubles them, and a purity above 100% is impossible.
The curve levels off at 240 cm3. 240 ÷ 24 000 = 0.010 mol CO2. The ratio CaCO3 : CO2 is 1 : 1, so 0.010 mol of CaCO3 (Mr 100) = 1.0 g. 0.44 g is the mass of carbon dioxide; 10 g divides 240 by 2400 instead of 24 000; 0.10 g is the result of an extra factor of 10 in the volume conversion.
Equal volumes of gases at the same temperature and pressure contain the same number of moles, so gas volumes react in the ratio of the equation. CO : O2 is 2 : 1, so 60 cm3 of CO needs 30 cm3 of O2. 60 cm3 ignores the ratio; 120 cm3 uses it upside down; 15 cm3 halves twice.
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.