Equations & reacting masses

5 practice questions

Balanced equations and state symbols

Learning goalBalance atoms using coefficients and show the physical states.

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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 symbol	Meaning
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Dissolved in water

Worked example

Aluminium reacts with oxygen

Balance Al + O₂ → Al₂O₃.

Worked answer

Use 2 Al₂O₃ to make 6 oxygen atoms on the right.
Use 3 O₂ to supply 6 oxygen atoms.
Use 4 Al to supply 4 aluminium atoms.

4Al(s) + 3O₂(g) → 2Al₂O₃(s).

EXAM TRAPA coefficient multiplies every atom in the following formula. An equation ratio is a mole ratio, not a mass ratio.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 17

(a) Balance Al + Cl₂ → AlCl₃, including state symbols for the solid metal, gaseous chlorine and solid product. (2 marks)

(b) Balance KClO₃ → KCl + O₂, including state symbols for the two solids and gaseous oxygen. (2 marks)

Total: 5 marks

Show answer & marking points

(a) 2Al(s) + 3Cl₂(g) → 2AlCl₃(s): coefficients [1]; states [1].

(b) 2KClO₃(s) → 2KCl(s) + 3O₂(g): coefficients [1]; states [1].

✓ DONE WHENBalance atoms using coefficients and show the physical states.

Ionic equations and spectator ions

Learning goalCancel unchanged aqueous ions while conserving both atoms and charge.

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WRITE IONIC FORMULAE FIRST

An ionic formula has zero total charge. Choose the smallest whole-number ratio of positive and negative ions; use brackets if more than one polyatomic ion is needed. For example, Zn²⁺ and CO₃²⁻ combine in a 1:1 ratio to give ZnCO₃. Recall the following ions; other ion charges are supplied when required.

Ion	Formula and charge
Nitrate	NO₃⁻
Carbonate	CO₃²⁻
Sulfate	SO₄²⁻
Hydroxide	OH⁻
Ammonium	NH₄⁺
Zinc	Zn²⁺
Silver	Ag⁺

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.

Worked example

A silver chloride precipitate

Mix aqueous silver nitrate and sodium chloride.

Write the balanced full equation and the ionic equation, including state symbols. Identify the spectator ions.

Worked answer

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

Cancel Na⁺ and NO₃⁻, the spectator ions.

Ag⁺(aq) + Cl⁻(aq) → AgCl(s).

EXAM TRAPDo not split the precipitate into ions. Check the total electrical charge as well as each element.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Exam-style question 18

Aqueous barium nitrate reacts with aqueous sodium carbonate to form solid barium carbonate and aqueous sodium nitrate. The barium ion is Ba²⁺.

(a) Write the balanced full equation, including state symbols. (2 marks)

(b) Write the ionic equation, including state symbols. (2 marks)

(d) Write the formulae of zinc hydroxide, silver sulfate and ammonium nitrate. (3 marks)

Total: 8 marks

Show answer & marking points

(a) Ba(NO₃)₂(aq) + Na₂CO₃(aq) → BaCO₃(s) + 2NaNO₃(aq): formulae and coefficients [1]; states [1].

(b) Ba²⁺(aq) + CO₃²⁻(aq) → BaCO₃(s): reacting species and balance [1]; states [1].

(d) Zn(OH)₂ [1]; Ag₂SO₄ [1]; NH₄NO₃. [1]

✓ DONE WHENCancel unchanged aqueous ions while conserving both atoms and charge.

Reacting masses and reverse calculations

Learning goalUse a balanced mole ratio between two substances.

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A reacting-mass calculation has one important bridge: the balanced equation connects moles of one substance to moles of another. It does not connect their gram values directly.

Write the balanced equation above your grid.
Put in the known mass and the relevant molar masses.
Use n = m/M to reach the moles row.
Move across that row using the equation's coefficient ratio.
Use m = nM to find the required mass.

The grid works in either direction. If the product mass is known, start there and follow the mole ratio backwards. Use g mol⁻¹ for M, which is numerically equal to the corresponding relative mass.

Worked example

Mass of aluminium oxide

5.40 g Al reacts with excess oxygen. Find the mass of Al₂O₃. M(Al) = 27.0; M(Al₂O₃) = 102.0 g mol⁻¹.

Worked answer

4Al + 3O₂ → 2Al₂O₃

Quantity	Al	Al₂O₃
Mass / g	5.40	m = n × M 0.100 × 102.0
M / g mol⁻¹	27.0	102.0
Amount / mol	n = m/M 5.40/27.0 = 0.200	n = n(known) × equation ratio 0.200 × 2/4 = 0.100

10.2 g Al₂O₃.

EXAM TRAPDo not multiply a mass by a coefficient ratio. Convert to moles first.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.

Quick check 19

Aluminium reacts with chlorine: 2Al + 3Cl₂ → 2AlCl₃.

(a) State the mole ratio n(Al) : n(Cl₂) : n(AlCl₃). [1 mark]

(b) Calculate the number of moles of AlCl₃ formed from 0.240 mol Cl₂ when aluminium is in excess. [2 marks]

Total: 4 marks

Show answer & marking points

(a)

n(Al) : n(Cl₂) : n(AlCl₃) = 2 : 3 : 2 [1]

(b)

n(AlCl₃) = n(known substance) × coefficient(target)/coefficient(known substance)
n(AlCl₃) = 0.240 × 2 ÷ 3 [1]

n(AlCl₃) = 0.160 mol [1]

(c)

Coefficients compare amounts in moles; particle masses differ, so they are not a mass ratio [1]

Exam-style question 20

Iron reacts with excess oxygen: 4Fe + 3O₂ → 2Fe₂O₃. Calculate the mass of Fe₂O₃ formed from 8.37 g Fe. M(Fe) = 55.8 and M(Fe₂O₃) = 159.6 g mol⁻¹. (3 marks)

Total: 3 marks

Show answer & marking points

n(Fe) = m/M
n(Fe) = 8.37/55.8 = 0.150 mol. [1] n(Fe₂O₃) = n(known substance) × coefficient(target)/coefficient(known substance)
n(Fe₂O₃) = 0.150 × 2/4 = 0.0750 mol. [1] m = n × M
m = 0.0750 × 159.6 = 11.97 g, or 12.0 g to 3 significant figures. [1]

Exam-style question 21

Sodium hydrogencarbonate decomposes: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂. Calculate the mass of NaHCO₃ needed to form 3.18 g Na₂CO₃. M(NaHCO₃) = 84.0 and M(Na₂CO₃) = 106.0 g mol⁻¹. (3 marks)

Total: 3 marks

Show answer & marking points

n(Na₂CO₃) = m/M
n(Na₂CO₃) = 3.18/106.0 = 0.0300 mol. [1] n(NaHCO₃) = n(known substance) × coefficient(target)/coefficient(known substance)
n(NaHCO₃) = 2 × 0.0300 = 0.0600 mol. [1] m = n × M
m = 0.0600 × 84.0 = 5.04 g. [1]

✓ DONE WHENUse a balanced mole ratio between two substances.

Equations & reacting masses: OCR A A-level Chemistry