Equations & reacting masses
Equations & reacting masses
Balance equations without changing formulae
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A reaction rearranges atoms; it does not create or destroy them. A balanced equation must therefore have the same number of atoms of each element on both sides.
- Write the correct formulae before balancing.
- Count the atoms on each side. Start with an element appearing in one reactant and one product if possible.
- Change the coefficients in front of whole formulae. A coefficient multiplies every atom in that formula.
- Recount every element, use the smallest whole-number coefficients and check the state symbols.
Never change a subscript to make the numbers balance. Changing H₂O to H₂O₂ would change the substance, not balance the original reaction.
| State | Meaning |
|---|---|
| (s) | Solid |
| (l) | Liquid |
| (g) | Gas |
| (aq) | Dissolved in water |
Aluminium burns in oxygen
Balance Al + O2 → Al2O3.
Worked answer
Two Al2O3 formula units contain six O atoms, so use three O2 molecules. This requires four Al atoms.
Neutralisation
Balance Ca(OH)2 + HCl → CaCl2 + H2O.
Worked answer
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
- Solid KClO₃ decomposes into solid KCl and oxygen gas. Write a balanced equation including state symbols. (2 marks)
- Balance Al(s) + HCl(aq) → AlCl₃(aq) + H₂(g). (2 marks)
- Balance Ca₃(PO₄)₂(s) + H₂SO₄(aq) → CaSO₄(s) + H₃PO₄(aq). (2 marks)
- Explain why changing a formula subscript is not a valid balancing method. (1 mark)
Total: 7 marks
Show answer & marking points
(a) 2KClO₃(s) → 2KCl(s) + 3O₂(g): coefficients [1]; states [1].
(b) 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g): Al/AlCl₃ coefficients [1]; HCl/H₂ coefficients [1].
(c) Ca₃(PO₄)₂(s) + 3H₂SO₄(aq) → 3CaSO₄(s) + 2H₃PO₄(aq): Ca/S species [1]; phosphate/acid coefficients [1].
(d) The subscript is part of the substance’s formula; changing it changes the substance. [1]
Write full and net ionic equations
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An ionic equation removes the ions that are only watching the reaction. These spectator ions appear unchanged on both sides.
- Start with the balanced full equation, including state symbols.
- Separate appropriate aqueous ionic substances, strong acids and strong alkalis into their ions. Keep the coefficients.
- Keep solids, liquids, gases and undissociated molecular substances intact. An (aq) label alone does not mean every substance should be split.
- Cancel spectator ions that are identical on both sides.
- Check both atom numbers and total charge.
For a strong acid reacting with a strong alkali, the reacting particles are H⁺(aq) and OH⁻(aq), which form water. The remaining ions do not need to appear in the net equation.
A silver chloride precipitate
Write the ionic equation for AgNO3(aq) + NaCl(aq).
Worked answer
Na+ and NO3− are spectator ions.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq).
- Write the full ionic equation, including state symbols. (3 marks)
- Identify both spectator ions. (2 marks)
- Write the net ionic equation, including state symbols. (2 marks)
- Explain why Cu(OH)₂ is not separated into aqueous ions. (1 mark)
- Write the net ionic equation for a strong acid reacting with a strong alkali, including states. (2 marks)
Total: 10 marks
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(a) Cu²⁺(aq) + SO₄²⁻(aq) + 2Na⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s) + 2Na⁺(aq) + SO₄²⁻(aq). Appropriate dissociation [1]; species/coefficients [1]; states and conserved charge [1].
(b) Na⁺ [1] and SO₄²⁻. [1]
(c) Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s): species/ratio [1]; states and charge [1].
(d) It is an insoluble solid, rather than freely separated aqueous ions. [1]
(e) H⁺(aq) + OH⁻(aq) → H₂O(l): species/stoichiometry [1]; states [1].
Calculate reacting masses with a grid
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A reacting-mass calculation has one important bridge: the balanced equation connects moles of one substance to moles of another.
The coefficients in a balanced equation compare moles, not masses. In the grid, work across the amount row using these coefficients before converting back to mass.
- Write the balanced equation above your grid.
- Put in the known mass and the relevant Aᵣ or Mr values, with no units on relative masses.
- Divide mass in grams by the relative mass to reach the moles row.
- Use the equation coefficients to find moles of the other substance.
- Multiply by its relative mass to find mass in grams.
The grid also works backwards from a product mass. Use Mr for molecules or formula units, and Aᵣ for atoms. Keep mass in grams and amount in moles.
Mr relative molecular or formula mass; no unit.
Magnesium to magnesium oxide
2Mg + O2 → 2MgO. Find the mass of MgO from 4.86 g Mg in excess O2. Ar(Mg) = 24.3; Mr(MgO) = 40.3.
Worked answer
| Quantity | Mg | O2 | MgO |
|---|---|---|---|
| Mass / g | 4.86 | In excess | 8.06 |
| Ar or Mr (no unit) | 24.3 | 32.0 | 40.3 |
| Amount / mol | n = m/Aᵣ 4.86/24.3 = 0.200 | Not needed | n = n(known) × equation ratio 0.200 × 2/2 = 0.200 |
m(MgO) = 0.200 × 40.3
Oxygen from peroxide
2H2O2 → 2H2O + O2. Find the mass of oxygen from 6.80 g H2O2. Mr(H2O2) = 34.0; Mr(O2) = 32.0.
Worked answer
| Quantity | H2O2 | O2 |
|---|---|---|
| Mass / g | 6.80 | 3.20 |
| Ar or Mr (no unit) | 34.0 | 32.0 |
| Amount / mol | n = m/Mᵣ 6.80/34.0 = 0.200 | n = n(known) × equation ratio 0.200 × 1/2 = 0.100 |
m(O2) = 0.100 × 32.0
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Sulfur dioxide reacts with oxygen: 2SO₂(g) + O₂(g) → 2SO₃(g).
- Calculate the amount in moles of SO₃ formed from 0.350 mol SO₂ with O₂ in excess. (1 mark)
- Calculate the amount in moles of O₂ needed for 0.350 mol SO₂. (1 mark)
- Calculate the amount in moles of SO₃ formed from 0.125 mol O₂ with SO₂ in excess. (1 mark)
- Explain why the coefficients cannot be applied directly to masses in grams. (1 mark)
Total: 4 marks
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(a) 0.350 mol SO₃. [1] (b) 0.350 × 1/2 = 0.175 mol O₂. [1] (c) 0.125 × 2 = 0.250 mol SO₃. [1]
(d) The coefficients give particle/amount ratios; different substances have different masses per mole. [1]
Titanium is formed in the following reaction: TiCl₄(g) + Mg(l) → Ti(s) + MgCl₂(l). A 7.29 g magnesium sample reacts with excess TiCl₄. Use Aᵣ(Mg) = 24.3 and Aᵣ(Ti) = 47.9.
- Balance the equation. (1 mark)
- Calculate the mass of titanium formed, to three significant figures. (3 marks)
Total: 4 marks
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(a) TiCl₄(g) + 2Mg(l) → Ti(s) + 2MgCl₂(l). [1]
(b) n(Mg) = m/Aᵣ
n(Mg) = 7.29/24.3 = 0.300 mol. [1] n(Ti) = n(known substance) × coefficient(target)/coefficient(known substance)
n(Ti) = 0.300/2 = 0.150 mol. [1] m(Ti) = n × Aᵣ
m(Ti) = 0.150 × 47.9 = 7.185 g = 7.19 g. [1]
Work backwards and calculate purity
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If a product mass is given, work back through the balanced equation to find how much reactant produced it. The route is the same as before, just travelled in the other direction.
For purity, distinguish the mass of pure reacting substance from the mass of the whole sample. Impurities contribute to the sample mass but, in the stated method, do not contribute to the measured reaction.
- Use the product data to find moles of the pure reactant.
- Convert those moles into the mass of pure reactant.
- Divide by the total sample mass and multiply by 100.
Iron oxide required
Fe₂O₃ + 3CO → 2Fe + 3CO₂. Calculate the mass of Fe₂O₃ required to produce 11.16 g Fe. Aᵣ(Fe) = 55.8; Mᵣ(Fe₂O₃) = 159.6.
Worked answer
| Quantity | Fe2O3 | Fe |
|---|---|---|
| Mass / g | 15.96 | 11.16 |
| Ar or Mr (no unit) | 159.6 | 55.8 |
| Amount / mol | n = n(known) × equation ratio 0.200 × 1/2 = 0.100 | n = m/Aᵣ 11.16/55.8 = 0.200 |
m(Fe2O3) = 0.100 × 159.6 = 15.96 g
Calcium carbonate purity
A 5.00 g impure CaCO3 sample gives 1.76 g CO2 with excess acid. CaCO3 + 2HCl → CaCl2 + CO2 + H2O. Mr(CO2) = 44.0; Mr(CaCO3) = 100.1.
Calculate the percentage purity by mass of the calcium carbonate sample. Assume the impurities do not react with the acid.
Worked answer
| Quantity | CaCO3 | CO2 |
|---|---|---|
| Mass / g | 4.004 pure | 1.76 |
| Ar or Mr (no unit) | 100.1 | 44.0 |
| Amount / mol | 0.0400 | n = m/Mᵣ 1.76/44.0 = 0.0400 |
Purity = 4.004 / 5.00 × 100
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g). Calculate the mass of Fe₂O₃ needed to produce 8.37 g Fe with CO in excess. Use Aᵣ(Fe) = 55.8 and Mᵣ(Fe₂O₃) = 159.6. (3 marks)
Total: 3 marks
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n(Fe) = m/Aᵣ
n(Fe) = 8.37/55.8 = 0.150 mol. [1] n(Fe₂O₃) = n(known substance) × coefficient(target)/coefficient(known substance)
n(Fe₂O₃) = 0.150/2 = 0.0750 mol. [1] m(Fe₂O₃) = n × Mᵣ
m(Fe₂O₃) = 0.0750 × 159.6 = 11.97 g = 12.0 g to three significant figures. [1]
12.5 g limestone containing CaCO₃ and inert impurities reacts with excess HCl, producing 4.40 g CO₂. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. Use Mᵣ(CO₂) = 44.0 and Mᵣ(CaCO₃) = 100.1.
- Calculate the percentage by mass of CaCO₃ in the limestone. (4 marks)
- Explain why HCl must be in excess. (1 mark)
Total: 5 marks
Show answer & marking points
Percentage purity = (mass of pure substance ÷ mass of sample) × 100
(a) n(CO₂) = m/Mᵣ
n(CO₂) = 4.40/44.0 = 0.100 mol. [1] The 1:1 ratio gives 0.100 mol CaCO₃. [1] Pure CaCO₃ mass = 0.100 × 100.1 = 10.01 g. [1] Purity = 10.01/12.5 × 100 = 80.1%. [1]
(b) Enough HCl must be available to react with all the CaCO₃, so the measured CO₂ represents the carbonate rather than a shortage of acid. [1]