A balanced equation is more than a description of a reaction — it's a precise numerical recipe. The coefficients tell you exactly how many moles of each substance react and form. Mastering mole ratios is the gateway to every calculation in Inquiry Question 1.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
If a recipe says "mix 2 cups of flour with 1 cup of sugar", you can scale it up or down, but the ratio must stay 2:1. How do you think chemists use a similar idea to figure out how much product they'll get from a chemical reaction? What information in the reaction equation would they use?
Type your initial thoughts below:
Record your ideas in your workbook.
📚 Core Content
Wrong: The limiting reagent is the one present in the smallest mass.
Right: The limiting reagent is the reactant that runs out first based on mole ratios, not mass.
In any chemical reaction, atoms are neither created nor destroyed — they are rearranged. This means the total mass of reactants always equals the total mass of products. It also means a chemical equation must have the same number of atoms of each element on both sides.
An unbalanced equation like H₂ + O₂ → H₂O is chemically incomplete. Balancing it — 2H₂ + O₂ → 2H₂O — makes the atom count equal on both sides and gives the equation quantitative meaning.
The inspection method works for most Year 11 equations. Follow these steps:
Once an equation is balanced, the coefficients give the mole ratio — the exact proportions in which substances react and are produced. This ratio holds at any scale.
If you know how many moles of one species you have, you can find moles of any other species using the ratio as a conversion factor:
🧮 Worked Examples
🧪 Activities
1 Balance: Mg + O₂ → MgO. Then state the mole ratio of Mg to MgO.
2 In N₂ + 3H₂ → 2NH₃ — if 6.00 mol of H₂ reacts, how many moles of NH₃ form?
3 Balance: Al + HCl → AlCl₃ + H₂. State the ratio of HCl to H₂.
4 In 4NH₃ + 5O₂ → 4NO + 6H₂O — if 0.800 mol of NH₃ reacts, how many moles of H₂O form?
5 Balance: C₂H₆ + O₂ → CO₂ + H₂O. Then find n(O₂) needed to burn 1.50 mol of C₂H₆.
Record your working for Q1–5:
Complete in workbook.
| Balanced Equation | Given | Find |
|---|---|---|
| 2Mg + O₂ → 2MgO | n(Mg) = 3.00 mol | n(MgO) = ? |
| 2Mg + O₂ → 2MgO | n(O₂) = 1.50 mol | n(MgO) = ? |
| N₂ + 3H₂ → 2NH₃ | n(N₂) = 2.00 mol | n(H₂) needed = ? |
| N₂ + 3H₂ → 2NH₃ | n(NH₃) = 5.00 mol | n(N₂) needed = ? |
| 4Fe + 3O₂ → 2Fe₂O₃ | n(Fe) = 1.20 mol | n(Fe₂O₃) = ? |
| 4Fe + 3O₂ → 2Fe₂O₃ | n(O₂) = 0.600 mol | n(Fe) needed = ? |
Show working for each row:
Complete table in workbook.
At the start of this lesson, you thought about how chemists use ratios from recipes to scale chemical reactions.
The answer: chemists use the coefficients in the balanced equation as mole ratios. Just as a 2:1 flour:sugar recipe scales proportionally, the equation coefficients tell you how many moles of each reactant and product are involved. For example, in N₂ + 3H₂ → 2NH₃, the ratio N₂:H₂:NH₃ = 1:3:2. Given any number of moles of one substance, you can calculate the others by multiplying by the ratio.
Reflect: how did your initial thinking compare to mole ratios?
Write a reflection in your workbook.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. (a) Balance the equation: C₃H₈ + O₂ → CO₂ + H₂O. (b) State the mole ratio of C₃H₈ to CO₂. (c) If 0.500 mol of C₃H₈ burns completely, how many moles of CO₂ and H₂O are produced? 4 MARKS
Show all steps:
Answer in workbook.
7. Explain the difference between a coefficient and a subscript in a chemical equation. Why can coefficients be changed when balancing but subscripts cannot? 3 MARKS
Write in full sentences:
Answer in workbook.
8. In the Haber process: N₂ + 3H₂ → 2NH₃. (a) How many moles of H₂ are required to produce 6.00 mol of NH₃? (b) If a factory produces 500 mol of NH₃ per hour, how many moles of N₂ are consumed per hour? 3 MARKS
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Answer in workbook.
9. Iron(III) oxide reacts with carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂. (a) If 0.400 mol of Fe₂O₃ reacts with excess CO, how many moles of Fe are produced? (b) If only 0.600 mol of Fe is obtained from this amount, calculate the percentage yield. (c) Identify one reason the actual yield might be less than the theoretical yield. 5 MARKS
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Answer in workbook.
10. Consider the reaction: 2SO₂ + O₂ → 2SO₃. A student claims: "If I start with 4.00 mol SO₂ and 4.00 mol O₂, the mole ratio says I need 2:1, so O₂ is in large excess and all SO₂ will react to give 4.00 mol SO₃." Evaluate this claim — is the student's reasoning about the limiting reagent correct? What is the maximum moles of SO₃ possible? 4 MARKS
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Answer in workbook.
1. C — 2:3. NH₃:H₂O = 4:6 = 2:3. Read coefficients directly from the balanced equation.
2. B. Check: Left: 4H, 2O. Right: 4H, 2O. ✓ Option A has 2H, 2O → 2H, 1O — not balanced. Options C and D change subscripts or split atoms — chemically invalid.
3. A — 8.00 mol. Ratio N₂:NH₃ = 1:2. n(NH₃) = 4.00 × 2 = 8.00 mol.
4. D. Subscripts define the chemical formula of a compound. Changing a subscript (FeCl₃ → FeCl₂) produces a different substance entirely. Only coefficients can be adjusted when balancing.
5. C — 2.25 mol. Ratio Al:Cl₂ = 2:3. n(Cl₂) = 1.50 × (3÷2) = 2.25 mol.
Q6 (4 marks):
(a) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O(b) Ratio C₃H₈ : CO₂ = 1 : 3
(c) n(CO₂) = 0.500 × 3 = 1.50 mol n(H₂O) = 0.500 × 4 = 2.00 molQ7 (3 marks): A coefficient is the large number written in front of a chemical formula in an equation (e.g. the 2 in 2H₂O). It represents the number of moles of that substance involved in the reaction. A subscript is the small number written within a formula (e.g. the 2 in H₂O) indicating the number of atoms of that element in one molecule. Coefficients can be changed because they simply scale the quantities of substances in the reaction without altering what those substances are. Subscripts cannot be changed because doing so would produce a completely different compound — for example, changing H₂O to H₂O₂ changes water to hydrogen peroxide.
Q8 (3 marks):
(a) Ratio H₂:NH₃ = 3:2; n(H₂) = 6.00 × (3÷2) = 9.00 mol (b) Ratio N₂:NH₃ = 1:2; n(N₂) = 500 × (1÷2) = 250 mol per hour6. C — The ratio H₂:H₂O = 2:2 = 1:1, so 3.0 mol H₂ → 3.0 mol H₂O ✓. O₂ needed = 3.0 ÷ 2 = 1.5 mol; we have 3.0 mol O₂ (excess). H₂ is limiting and 3.0 mol H₂O is correct numerically. However the student's reasoning about the "1:1 ratio" is misleading — the actual coefficients are 2:2, simplified to 1:1. Answer C correctly identifies the reasoning flaw.
7. B — n(Al) = 54.0 ÷ 26.982 = 2.001 mol. n(Fe₂O₃) = 80.0 ÷ 159.69 = 0.5009 mol. Ratio 2:1 → need 2 × 0.5009 = 1.002 mol Al. We have 2.001 mol Al > 1.002 mol required → Fe₂O₃ is limiting. n(Fe) = 2 × 0.5009 = 1.002 mol. m(Fe) = 1.002 × 55.845 = 55.9 g.
Q9 (5 marks):
(a) n(Fe) = 0.400 × (2÷1) = 0.800 mol [1] (b) % yield = (0.600 ÷ 0.800) × 100 = 75.0% [1](c) Possible reasons: incomplete reaction (not all Fe₂O₃ converted); side reactions producing other products; loss of product during transfer/filtration; impure reactants [1 for any valid reason].
Q10 (4 marks):
O₂ required for 4.00 mol SO₂ = 4.00 × (1÷2) = 2.00 mol [1]. We have 4.00 mol O₂ — this is double what's needed, so O₂ is indeed in excess and SO₂ is the limiting reagent [1]. n(SO₃) = 4.00 × (2÷2) = 4.00 mol [1]. The student's conclusion (4.00 mol SO₃) is correct [1]. Their reasoning that O₂ is "in large excess" is also correct (2.00 mol O₂ reacts; 2.00 mol O₂ remains).
Climb platforms, hit checkpoints, and answer questions on mole ratios from balanced equations and stoichiometric calculations. Quick recall from lessons 1–11.
Tick when you've finished all activities and checked your answers.