Balanced equations and reacting massesEdexcel A-Level Chemistry: Revision notes
Section 1
Writing balanced equations
A balanced equation has the same number of each type of atom on both sides, and the same total charge. Use state symbols: (s), (l), (g), (aq).
- Balance by changing the numbers in front of formulae, never the formulae themselves.
- Soluble ionic compounds in solution are (aq); precipitates are (s); gases are (g).
Example: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
Check the charge as well as the atoms when you write an ionic equation.
Section 2
Ionic equations
An ionic equation shows only the species that change. Spectator ions are present unchanged on both sides and are cancelled.
Precipitation: Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) becomes Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s), with K⁺ and NO₃⁻ as spectators.
Acid and carbonate: CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l).
Solids, gases and liquids stay as full formulae; only dissolved ionic compounds are split into ions.
Splitting an insoluble solid such as CaCO₃ into ions. Only dissolved ionic species are written as ions.
Section 3
Amounts: mass, gas volume and solutions
Convert everything to moles, then use the equation ratio.
- Mass: n = m ÷ M
- Solutions: n = c × V, with V in dm³ (divide cm³ by 1000)
- Gases at room temperature and pressure: n = V ÷ 24.0, with V in dm³ (molar volume 24.0 dm³ mol⁻¹)
The coefficients in a balanced equation give the mole ratio.
Worked example: 2.50 g CaCO₃ with excess acid: n = 2.50 ÷ 100.1 = 0.0250 mol; 1 : 1 with CO₂, so V = 0.0250 × 24.0 = 0.600 dm³.
Section 4
Reacting masses
To find a reacting mass from an equation:
- Convert the known mass to moles.
- Use the equation ratio to find moles of the other substance.
- Convert back to mass using M.
Example: 460 g CaCO₃ (4.60 mol) in CaCO₃ → CaO + CO₂ gives 4.60 mol CaO, mass 4.60 × 56.1 = 258 g.
If one reactant is in excess, the limiting reactant decides the amount of product. Work from the limiting reactant. For impure samples, find the mass of pure compound first (mass × percentage).
Using the mass of the excess reactant to find the amount of product. Always use the limiting reactant.
Section 5
Equations and observations
Link the equation to what you see:
- Displacement: Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s): blue solution fades, red-brown copper deposits.
- Acid + carbonate: effervescence of CO₂.
- Acid + metal: effervescence of hydrogen, Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).
- Neutralisation: H⁺(aq) + OH⁻(aq) → H₂O(l), with a temperature rise.
- Precipitation: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s): yellow solid; Ag⁺(aq) + Cl⁻(aq) → AgCl(s): white solid.
Must know
- Balance atoms and charge; include state symbols
- Ionic equations: cancel spectator ions, keep solids, liquids and gases whole
- n = m ÷ M, n = cV (dm³), n = V ÷ 24.0 for gases at RTP
- Use the mole ratio from the equation; work from the limiting reactant
- Link observations to the equation: colour change, precipitate, effervescence
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Balanced equations and reacting masses
- A student reacts 2.50 g of marble chips (calcium carbonate) with excess dilute hydrochloric acid in a conical flask.Calculate the volume of carbon dioxide produced at room temperature and pressure. M(CaCO₃) = 100.1 g mol⁻¹ and the molar volume of a gas is 24.0 dm³ mol⁻¹.2 marks
- A student mixes 20.0 cm³ of 0.150 mol dm⁻³ aqueous lead(II) nitrate with excess aqueous potassium iodide. A precipitate forms.Calculate the maximum mass of lead(II) iodide formed. M(PbI₂) = 461.0 g mol⁻¹.2 marks
- A student adds 1.20 g of magnesium ribbon to 50.0 cm³ of 0.500 mol dm⁻³ aqueous copper(II) sulfate and stirs. The mixture warms up.Write an ionic equation, with state symbols, for the reaction and explain the observations in terms of this equation.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).