Balancing Equations and Stoichiometric Calculations Revision Notes | A Level AQA Chemistry
Write the correct formulae, then balance one element at a time, leaving hydrogen and oxygen until last. Free AQA notes and 14 questions answered.
Updated 5 Sept 2026 · Revise Right Team
Balancing Equations and Stoichiometric Calculations is part of Amount of substance in A Level AQA Chemistry.
5 min
Reading time
68
Lessons in this topic
A Level AQA
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
Write the correct formulae, then balance one element at a time, leaving hydrogen and oxygen until last. Free AQA notes and 14 questions answered.
Write and balance full and ionic equations for familiar and unfamiliar reactions, including correct state symbols
Use balanced equations to calculate masses, volumes of gases, and concentrations and volumes for reactions in solution
Apply stoichiometric reasoning to multi-step mole calculations with unit precision
Describe and evaluate the procedure for making up a volumetric solution and carrying out an acid-base titration
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Use this lesson to stay inside Amount of substance while connecting the detail back to the wider Physical chemistry topic.
Stoichiometry: The quantitative relationship between reactants and products in a chemical reaction, determined by the mole ratios in a balanced equation.
Molar volume of a gas: The volume occupied by one mole of any gas at room temperature and pressure (RTP), equal to 24 dm³ mol⁻¹.
Concentration: The amount of solute (in mol) dissolved per unit volume of solution (in dm³), expressed in mol dm⁻³.
Ionic equation: An equation showing only the species that change during a reaction, with spectator ions omitted, written with correct charges and state symbols.
Revision notes
Read the core explanations from Balancing Equations and Stoichiometric Calculations before testing yourself from memory.
1. Why Balancing Equations Is the Foundation of Everything
Every quantitative calculation in chemistry — masses, volumes, concentrations — starts with one thing: a balanced equation. Get the equation wrong and every number that follows is wrong too. This lesson builds the skills to write, balance, and use equations with the precision AQA examiners expect.
Think of a balanced equation as a recipe. The coefficients tell you the exact mole ratios of every ingredient and product. If a recipe calls for 2 eggs per 100 g of flour, you cannot use 3 eggs and expect the same result — chemistry works the same way.
2. Writing and Balancing Equations
A balanced equation satisfies two conditions: the number of atoms of each element is equal on both sides, and the total charge is equal on both sides (critical for ionic equations). State symbols — (s), (l), (g), (aq) — are required in AQA answers and carry marks.
For unfamiliar reactions, AQA will give you the reactants and products. Your job is to write the correct formulae and then balance. Work through one element at a time, leaving hydrogen and oxygen until last — they appear in the most places and are straightforward to adjust at the end.
Worked example: Balancing an Unfamiliar Equation
Balance the equation for the reaction of iron(III) oxide with carbon monoxide to form iron and carbon dioxide. Include state symbols.
$\text{Fe}_2\text{O}_3\text{(s) + CO(g)} \rightarrow \text{Fe(s) + CO}_2\text{(g)}$ — Iron(III) oxide is Fe₂O₃. Carbon monoxide is CO. Check each formula is correct before balancing.
Left: Fe = 2, O = 3 + 1 = 4, C = 1
Right: Fe = 1, O = 2, C = 1 — Fe and O are unbalanced. Tackle Fe first — there are 2 Fe on the left, so place a 2 in front of Fe on the right.
$\text{Fe}_2\text{O}_3\text{(s) + CO(g)} \rightarrow \text{2Fe(s) + CO}_2\text{(g)}$ — Fe is now balanced: 2 on each side.
Left O: 3 (from Fe₂O₃) + 1 per CO
Right O: 2 per CO₂
Need 3CO on left → 3CO₂ on right
$\text{Fe}_2\text{O}_3\text{(s) + 3CO(g)} \rightarrow \text{2Fe(s) + 3CO}_2\text{(g)}$ — 3 CO provides 3 extra O atoms, giving 3 + 3 = 6 O on the left. Right side: 3 × 2 = 6 O. Carbon: 3 on each side. Balanced.
Left: Fe = 2, O = 3 + 3 = 6, C = 3
Right: Fe = 2, O = 6, C = 3 — All atoms balance. State symbols are already included. The equation is complete.
Ionic equations strip out the spectator ions — the ions that are present but do not change during the reaction. They reveal the actual chemical change taking place and are frequently tested in AQA papers.
The method has three stages: write the full equation, split all soluble ionic compounds and strong acids into their ions, then cancel any species that appear identically on both sides. What remains is the net ionic equation.
Worked example: Deriving an Ionic Equation
Write the ionic equation for the reaction between sodium carbonate solution and dilute hydrochloric acid.
$\text{2Na}^+\text{(aq) + CO}_3^{2-}\text{(aq) + 2H}^+\text{(aq) + 2Cl}^-\text{(aq)} \rightarrow \text{2Na}^+\text{(aq) + 2Cl}^-\text{(aq) + H}_2\text{O(l) + CO}_2\text{(g)}$ — Soluble salts and strong acids split into ions. Water (l) and CO₂ (g) stay as molecules — they are not split.
Cancel 2Na⁺(aq) and 2Cl⁻(aq) from both sides. — These ions appear unchanged on both sides — they are spectators.
$\text{CO}_3^{2-}\text{(aq) + 2H}^+\text{(aq)} \rightarrow \text{H}_2\text{O(l) + CO}_2\text{(g)}$ — This is the ionic equation. Check charge balance: left = 2− + 2+ = 0; right = 0. Balanced in charge and atoms.
With a balanced equation in hand, you can calculate the mass of any reactant or product from the mass of any other. The mole is the bridge: convert mass to moles, apply the mole ratio from the equation, then convert back to mass.
Worked example: Mass Calculation from a Balanced Equation
Calculate the mass of calcium oxide produced when 5.00 g of calcium carbonate is completely decomposed. $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s) + CO}_2\text{(g)}$
$M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 40.1 + 12.0 + 48.0 = 100.1 \text{ g mol}^{-1}$ — Add the relative atomic masses of Ca, C, and three O atoms.
$n(\text{CaCO}_3) = \frac{5.00}{100.1} = 0.04995 \text{ mol}$ — Use n = m ÷ Mᵣ. Keep extra significant figures at this stage to avoid rounding errors.
Ratio CaCO₃: CaO = 1: 1
$n(\text{CaO}) = 0.04995 \text{ mol}$ — The equation shows 1 mol CaCO₃ produces 1 mol CaO, so the moles are equal.
$M_r(\text{CaO}) = 40.1 + 16.0 = 56.1 \text{ g mol}^{-1}$ — Calcium plus oxygen.
$m(\text{CaO}) = 0.04995 \times 56.1 = 2.80 \text{ g}$ — Use m = n × Mᵣ. Round to 3 significant figures to match the data given.
Answer: Mass of CaO = 2.80 g
4. Volumes of Gases
Gases behave differently from solids and liquids — at the same temperature and pressure, equal volumes of any gas contain equal numbers of moles. At room temperature and pressure (RTP, defined as 25 °C and 100 kPa), one mole of any gas occupies 24 dm³.
A common unit trap: if a volume is given in cm³, convert to dm³ by dividing by 1000 before using this formula. Leaving the volume in cm³ and dividing by 24 gives an answer that is 1000 times too small because the formula expects dm³, not cm³.
Worked example: Gas Volume Calculation
Calculate the volume of carbon dioxide gas produced at RTP when 6.00 g of calcium carbonate reacts completely with excess hydrochloric acid. $\text{CaCO}_3\text{(s) + 2HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq) + H}_2\text{O(l) + CO}_2\text{(g)}$
$M_r(\text{CaCO}_3) = 100.1 \text{ g mol}^{-1}$
$n(\text{CaCO}_3) = \frac{6.00}{100.1} = 0.05994 \text{ mol}$ — Use n = m ÷ Mᵣ.
Ratio CaCO₃: CO₂ = 1: 1
$n(\text{CO}_2) = 0.05994 \text{ mol}$ — One mole of CaCO₃ produces one mole of CO₂.
$V(\text{CO}_2) = 0.05994 \times 24 = 1.44 \text{ dm}^3$ — Multiply moles by the molar gas volume at RTP (24 dm³ mol⁻¹). Round to 3 significant figures.
Answer: Volume of CO₂ = 1.44 dm³ at RTP
5. Concentrations and Volumes in Solution
When reactions happen in solution, you need to connect concentration and volume to moles. Concentration (in mol dm⁻³) tells you how many moles are packed into each cubic decimetre of solution.
Worked example: Concentration Calculation Using a Balanced Equation
25.0 cm³ of sodium hydroxide solution of unknown concentration is neutralised by 20.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid. Calculate the concentration of the sodium hydroxide solution. $\text{NaOH(aq) + HCl(aq)} \rightarrow \text{NaCl(aq) + H}_2\text{O(l)}$
$V(\text{HCl}) = \frac{20.0}{1000} = 0.0200 \text{ dm}^3$
$V(\text{NaOH}) = \frac{25.0}{1000} = 0.0250 \text{ dm}^3$ — Always convert cm³ to dm³ before using n = c × V.
$n(\text{HCl}) = 0.100 \times 0.0200 = 0.00200 \text{ mol}$ — Use n = c × V with the known concentration and converted volume.
Ratio NaOH: HCl = 1: 1
$n(\text{NaOH}) = 0.00200 \text{ mol}$ — The balanced equation shows a 1:1 mole ratio, so moles of NaOH equal moles of HCl at the equivalence point.
$c(\text{NaOH}) = \frac{n}{V} = \frac{0.00200}{0.0250} = 0.0800 \text{ mol dm}^{-3}$ — Rearrange n = c × V to give c = n ÷ V. Units: mol ÷ dm³ = mol dm⁻³.
Answer: Concentration of NaOH = 0.0800 mol dm⁻³
6. Making a Volumetric Solution and Acid-Base Titration
Volume given
Correct conversion to dm3
Effect if you forget the conversion
24.0 cm3
0.0240 dm3
moles from n = cV or n = V/24 become 1000 times too small
250 cm3
0.250 dm3
concentration or moles are off by a factor of 1000
1.44 dm3
already in dm3
no conversion needed
The same stoichiometric logic appears in every AQA amount-of-substance task: identify the balanced equation, convert the given quantity into moles with the correct units, apply the coefficient ratio, then convert to the final quantity. The practical skill is deciding which quantity is actually known and making sure units such as cm³ and dm³ are handled before the calculation starts.
The acid-base titration is a required practical for AQA A Level Chemistry. It links directly to everything covered so far: you prepare a standard solution of known concentration, then use it to find the concentration of an unknown solution. The precision of your technique determines the quality of your result.
Notice how the titration calculation is exactly the same structure as the worked example in Section 5 — moles of known solution, mole ratio, then concentration of unknown. The practical skill and the calculation skill are inseparable.
7. Exam Technique and Common Mistakes
AQA mark schemes for stoichiometry questions follow a consistent pattern: one mark for the correct mole calculation, one mark for applying the mole ratio, and one mark for the final answer with correct units. Losing the ratio mark is the most common way students drop points — always write the ratio explicitly in your working.
Every calculation in this lesson — mass, gas volume, concentration — follows the same three-step logic: convert to moles, apply the mole ratio from the balanced equation, then convert to the required quantity. Master that pattern and stoichiometry becomes a reliable, repeatable process rather than a collection of disconnected formulae.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-05.
Write the correct formulae, then balance one element at a time, leaving hydrogen and oxygen until last, and finally check every atom and add state symbols. For example, Fe₂O₃(s) + CO(g) → Fe(s) + CO₂(g) balances as Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g), with 2 Fe, 3 C and 6 O on each side.
How do you calculate an unknown concentration from a titration?
Find the moles of the known solution using n = c × V with the volume in dm³, apply the mole ratio, then use c = n ÷ V for the unknown. If 20.0 cm³ of 0.100 mol dm⁻³ HCl neutralises 25.0 cm³ of NaOH, n = 0.00200 mol, so the concentration of NaOH = 0.00200 ÷ 0.0250 = 0.0800 mol dm⁻³.
How do you write an ionic equation?
Write the full balanced equation, split soluble ionic compounds and strong acids into their ions, then cancel the spectator ions that appear unchanged on both sides. For sodium carbonate and hydrochloric acid, the spectator ions Na⁺ and Cl⁻ cancel, leaving CO₃²⁻(aq) + 2H⁺(aq) → H₂O(l) + CO₂(g), which balances in both atoms and charge.
What is a spectator ion?
A spectator ion is an ion that is present in a reaction but appears unchanged on both sides of the equation. In the reaction of sodium carbonate with hydrochloric acid, Na⁺ and Cl⁻ are spectator ions, so they cancel to give the net ionic equation.
What is the formula linking moles, concentration and volume?
The formula is n = c × V, where n is moles, c is concentration in mol dm⁻³ and V is volume in dm³. Rearranging gives c = n ÷ V. For 20.0 cm³ of 0.100 mol dm⁻³ HCl, n = 0.100 × 0.0200 = 0.00200 mol.
What is the molar gas volume at room temperature and pressure?
At room temperature and pressure, defined as 25 °C and 100 kPa, one mole of any gas occupies 24 dm³. Equal volumes of any gas at the same temperature and pressure contain equal numbers of moles.
How do you calculate the mass of a product from a balanced equation?
Convert the known mass to moles using moles = mass ÷ molar mass, apply the mole ratio from the balanced equation, then convert back to mass. When 5.00 g of CaCO₃ decomposes, moles = 5.00 ÷ 100.1 = 0.04995 mol, giving 0.04995 mol of CaO, and 0.04995 × 56.1 = 2.80 g of CaO.
What are state symbols in chemical equations?
State symbols show the state of each species: (s) for solid, (l) for liquid, (g) for gas and (aq) for a solution in water. They are required in balanced equations, for example 2Mg(s) + O₂(g) → 2MgO(s).
How do you calculate the volume of a gas at room temperature and pressure?
At room temperature and pressure (25 °C and 100 kPa), one mole of any gas occupies 24 dm³, so volume = moles × 24. For example, 6.00 g of CaCO₃ reacting with excess acid gives 0.05994 mol of CO₂, which occupies 0.05994 × 24 = 1.44 dm³. Convert any volume in cm³ to dm³ by dividing by 1000.
What is the ionic equation for carbonate and acid?
The ionic equation for a carbonate reacting with an acid is CO₃²⁻(aq) + 2H⁺(aq) → H₂O(l) + CO₂(g). The charge balances: 2− + 2+ = 0 on the left and 0 on the right, and the spectator ions have gone.
How do you balance the equation for iron(III) oxide and carbon monoxide?
The balanced equation is Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g). Iron is balanced first with a 2 in front of Fe, then 3CO and 3CO₂ balance oxygen and carbon, giving 6 oxygen atoms and 3 carbon atoms on each side.
What is the difference between a full equation and an ionic equation?
A full equation shows all reactants and products with complete formulae, keeping soluble ionic compounds as full formulae. An ionic equation splits soluble ionic compounds and strong acids into ions and cancels spectator ions, showing only the actual chemical change.
How much calcium oxide forms from 5.00 g of calcium carbonate?
Decomposing 5.00 g of calcium carbonate produces 2.80 g of calcium oxide. Moles of CaCO₃ = 5.00 ÷ 100.1 = 0.04995 mol; the ratio CaCO₃ : CaO is 1 : 1; and mass of CaO = 0.04995 × 56.1 = 2.80 g.
What are the three steps in every stoichiometry calculation?
Every stoichiometry calculation follows three steps: convert the given quantity to moles, apply the mole ratio from the balanced equation, then convert to the required quantity. This works for masses, gas volumes and concentrations alike.