Stoichiometry — Mole Ratios
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.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
Key Relationships — L11
Coefficients in balanced equation = mole ratio
n(wanted) = n(given) × [coeff(wanted) ÷ coeff(given)]
Key facts
- Law of conservation of mass — atoms are neither created nor destroyed
- Balancing means equal atoms of each element on both sides
- Coefficients define the mole ratio of all species in a reaction
- Mole ratio is always taken from the balanced equation
Concepts
- Balance any equation by inspection
- Extract the mole ratio between any two species in a reaction
- Use ratios to find moles of one species given moles of another
Skills
- TODO
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.
Step 2. Count atoms of each element on each side.
Step 3. Add coefficients in front of formulae to make counts equal — never change subscripts.
Step 4. Start with the most complex molecule, balance metals and non-metals before hydrogen and oxygen, and balance oxygen last.
Step 5. Check all atoms balance and coefficients are in the simplest whole-number ratio.
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:
Example: Given 4 mol H₂, how many mol H₂O forms?
n(H₂O) = 4 × (2 ÷ 2) = 4 mol H₂O
Example: Given 3 mol O₂, how many mol H₂O forms?
n(H₂O) = 3 × (2 ÷ 1) = 6 mol H₂O
Law of Conservation of Mass: atoms are neither created nor destroyed; total mass of reactants equals total mass of products. Balanced equation coefficients give the mole ratio. To convert moles of one species to another: n(wanted) = n(given) × coeff(wanted) ÷ coeff(given).
Pause — copy the highlighted law and formula into your book before moving on.
Match each term on the left to its meaning on the right.
- Coefficient
- Subscript
- Mole ratio
- Conservation of mass
- The small number inside a formula — gives atoms per molecule and must never be changed when balancing.
- Atoms are rearranged, not created or destroyed, so total mass of reactants = total mass of products.
- The large number in front of a formula — the only thing you may change to balance an equation.
- The proportion in which species react and are produced, read directly from the coefficients of a balanced equation.
Worked examples · reveal as you go
Balance the equation: Fe + O₂ → Fe₂O₃
In 2H₂ + O₂ → 2H₂O — if 4.00 mol of H₂ reacts completely, how many moles of H₂O form?
In 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O — what mole ratio of HCl to Ca(OH)₂ is required for complete reaction? If 0.300 mol of Ca(OH)₂ is used, how many moles of HCl are needed?
Propane burns: C₃H₈ + O₂ → CO₂ + H₂O. (a) Balance the equation. (b) If 2.00 mol of C₃H₈ burns completely, how many moles of CO₂ and H₂O are produced?
Click two steps to swap them. Get the inspection method in the right order before you start balancing: C₃H₈ + O₂ → CO₂ + H₂O, then find n(CO₂) for 2.00 mol of C₃H₈.
- Balance hydrogen next: 8 H on the left → 4H₂O on the right.
- Write the unbalanced equation and count atoms of each element on each side.
- Apply the mole ratio C₃H₈ : CO₂ = 1 : 3 → n(CO₂) = 2.00 × (3 ÷ 1) = 6.00 mol.
- Balance carbon first: 3 C on the left → 3CO₂ on the right.
- Balance oxygen last: 3(2) + 4(1) = 10 O atoms → 5O₂. Balanced equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
Common errors · the 3 traps that cost marks
Using subscripts as the mole ratio instead of coefficients
In 2H₂O, the subscript 2 in H₂ means two hydrogen atoms per molecule — it is not the mole ratio. The coefficient 2 (in front of H₂O) is the mole ratio. Students who confuse these extract ratios like H:O = 2:1 when the correct ratio from 2H₂ + O₂ → 2H₂O is H₂:O₂ = 2:1. This gives a completely wrong answer.
✓ Fix: Only read coefficients (the numbers in front of formulae) for mole ratios. Never use subscripts. If a coefficient is missing (implied 1), write it explicitly before extracting the ratio.
Changing subscripts when trying to balance
H₂O₂ and H₂O are different compounds — you cannot change a subscript to balance an equation. Students sometimes write H₂O₃ or HO to make atoms balance. This is chemically wrong. You may only change or add coefficients in front of whole formulae.
✓ Fix: Treat all formulae as fixed. Only ever add or change the large numbers in front of each formula (the coefficients). Never touch the small numbers inside formulae (subscripts).
Assuming all reactions have a 1:1 mole ratio
Many students skip writing and checking the balanced equation, assume all ratios are 1:1, and get the wrong answer. In 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O, HCl reacts with Ca(OH)₂ in a 2:1 ratio — missing this doubles the error. This lesson has no calculation shortcuts. Always write the balanced equation first.
✓ Fix: Every stoichiometry problem — even ones that seem simple — must start with a balanced equation. Write it, extract the ratio explicitly, then calculate. No skipping steps.
Quick-fire practice · 5 reps +2 XP per reveal
For N₂ + 3H₂ → 2NH₃, how many moles of NH₃ form from 6.0 mol of H₂ (N₂ in excess)?
For 2Al + 3Cl₂ → 2AlCl₃, how many moles of Cl₂ react with 0.40 mol of Al?
In 2H₂ + O₂ → 2H₂O you have 3.0 mol H₂ and 2.0 mol O₂. Which reactant is limiting?
For CaCO₃ → CaO + CO₂, how many moles of CO₂ are released from 0.25 mol of CaCO₃?
For C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, how many moles of O₂ burn 0.10 mol of propane?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Complete the working below for the reaction N₂ + 3H₂ → 2NH₃. Given 6.00 mol of H₂, find n(NH₃). Type each missing number, then click Check.
Mole ratio H₂ : NH₃ = 3 :
Apply ratio: n(NH₃) = n(H₂) × coeff(NH₃) ÷ coeff(H₂)
n(NH₃) = 6.00 × ÷ = mol
Q1. 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?
Q2. 7. Explain the difference between a coefficient and a subscript in a chemical equation. Why can coefficients be changed when balancing but subscripts cannot?
Q3. 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?
Q4. 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.
Q5. 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?
📖 Comprehensive answers (click to reveal)
Activity 2 — Data Table Answers
Row 1: n(MgO) = 3.00 × (2÷2) = 3.00 mol Row 2: n(MgO) = 1.50 × (2÷1) = 3.00 mol Row 3: n(H₂) = 2.00 × (3÷1) = 6.00 mol Row 4: n(N₂) = 5.00 × (1÷2) = 2.50 mol Row 5: n(Fe₂O₃) = 1.20 × (2÷4) = 0.600 mol Row 6: n(Fe) = 0.600 × (4÷3) = 0.800 mol❓ Multiple Choice
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.
Short Answer Model Answers
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 hour❓ MC — Q6 & Q7
6. 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.
Short Answer — Q9 & Q10
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).
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