The problems are on the balancing equations worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A formula shows the ratio of atoms or ions in a substance, and for ionic compounds the charges must cancel. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on balancing equations, from formulae from ions to heating limestone: each solution shows the formula, the substitution and the final answer with its unit.
Compare each answer with your own and check the method as well as the final equation. Whenever an equation is balanced, count the atoms on both sides again, and for ionic equations check that the total charge is equal on both sides.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: HS-PS1-7
Problem 1 Formulae from ions
1. Na⁺ and Cl⁻ have equal and opposite charges: NaCl.2. Mg²⁺ and O²⁻: MgO.
3. Ca²⁺ needs two Cl⁻: CaCl₂.
NaCl, MgO, CaCl₂.
Problem 2 State symbols
1. (aq) means the substance is dissolved in water.2. Zinc is solid, hydrochloric acid and zinc chloride are in solution, and hydrogen is a gas.
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
Problem 3 Counting atoms
1. 2H₂SO₄: H 2 × 2 = 4, S 2, O 2 × 4 = 8.2. 3Mg(NO₃)₂: Mg 3, N 3 × 2 = 6, O 3 × 2 × 3 = 18.
3. Al₂(SO₄)₃: Al 2, S 3, O 3 × 4 = 12.
H 4, S 2, O 8; Mg 3, N 6, O 18; Al 2, S 3, O 12.
Problem 4 Balance the simple ones
1. Two Cl atoms on the left need 2HCl, and then H is 2 = 2.2. Two Cl need 2NaCl, so 2Na.
3. Two O need 2MgO, so 2Mg.
H₂ + Cl₂ → 2HCl; 2Na + Cl₂ → 2NaCl; 2Mg + O₂ → 2MgO
Problem 5 Reading particles
1. Left: N₂ has 2 N atoms and 3 H₂ have 6 H atoms. Right: 2 NH₃ have 2 N atoms and 6 H atoms.2. One N₂ and three H₂ form two NH₃.
N₂ + 3H₂ → 2NH₃
Problem 6 Word equation to symbols
1. Calcium carbonate CaCO₃, calcium oxide CaO and carbon dioxide CO₂.2. The equation is already balanced (Ca 1, C 1, O 3 on each side).
CaCO₃(s) → CaO(s) + CO₂(g)
Problem 7 The wrong subscript
1. H₂O₂ is hydrogen peroxide, a different substance from water. Subscripts must never be changed to balance an equation.2. Put the coefficient 2 in front of H₂O to give 2 O atoms, then 2 in front of H₂ to give 4 H atoms.
2H₂ + O₂ → 2H₂O
Problem 8 Formulae with brackets
1. Ca²⁺ and OH⁻: Ca(OH)₂.2. Al³⁺ and NO₃⁻: Al(NO₃)₃.
3. NH₄⁺ and SO₄²⁻: (NH₄)₂SO₄.
4. Fe³⁺ and O²⁻: Fe₂O₃ (6+ and 6−).
5. Cu²⁺ and NO₃⁻: Cu(NO₃)₂.
Ca(OH)₂, Al(NO₃)₃, (NH₄)₂SO₄, Fe₂O₃, Cu(NO₃)₂
Problem 9 Burning fuels
1. The diagram shows one CH₄, two O₂, one CO₂ and two H₂O: CH₄ + 2O₂ → CO₂ + 2H₂O.2. Carbon: 3CO₂. Hydrogen: 4H₂O. Oxygen: 6 + 4 = 10, so 5O₂.
3. Carbon: 2CO₂. Hydrogen: 3H₂O. Oxygen on the right: 4 + 3 = 7; ethanol supplies 1, so 3O₂ supply the other 6.
CH₄ + 2O₂ → CO₂ + 2H₂O; C₃H₈ + 5O₂ → 3CO₂ + 4H₂O; C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
Problem 10 Balance the metals
1. Oxygen 6 as the common number: 3O₂ and 2Fe₂O₃, then 4Fe.2. Chlorine 3 on the right needs 3HCl for one Al; double all: 2Al + 6HCl → 2AlCl₃ + 3H₂ (H 6 = 6).
3. Hydrogen on the right is 3 (NaOH 1 and H₂ 2); use 2Na + 2H₂O → 2NaOH + H₂ (H 4 = 4, O 2 = 2).
4Fe + 3O₂ → 2Fe₂O₃; 2Al + 6HCl → 2AlCl₃ + 3H₂; 2Na + 2H₂O → 2NaOH + H₂
Problem 11 Neutralisation equations
1. H₂SO₄ has 2 H⁺, so it needs 2 NaOH: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l).2. Ca(OH)₂ has 2 OH⁻, so it needs 2 HCl: 2HCl(aq) + Ca(OH)₂(aq) → CaCl₂(aq) + 2H₂O(l).
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l); 2HCl(aq) + Ca(OH)₂(aq) → CaCl₂(aq) + 2H₂O(l)
Problem 12 A precipitate forms
1. AgNO₃(aq) + KCl(aq) → AgCl(s) + KNO₃(aq).2. K⁺ and NO₃⁻ are in solution before and after, so they are the spectators.
3. Silver and chloride ions form the solid.
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Problem 13 Which formulae are wrong?
1. NaO, MgCl and CaNO₃ are wrong. Al₂O₃ and K₂SO₄ are correct.2. Na₂O, MgCl₂ and Ca(NO₃)₂.
3. Na⁺ has one positive charge, but O²⁻ has two negative charges, so two sodium ions are needed.
Na₂O, MgCl₂, Ca(NO₃)₂
Problem 14 Conservation of mass
1. 4.8 + 3.2 = 8.0 g.2. The law of conservation of mass: the total mass of reactants equals the total mass of products.
3. Oxygen from the air joined the magnesium, so the solid gains the mass of the oxygen.
8.0 g of magnesium oxide.
Problem 15 Zinc in copper sulfate
1. Zn(s) + Cu²⁺(aq) + SO₄²⁻(aq) → Zn²⁺(aq) + SO₄²⁻(aq) + Cu(s).2. Sulfate, SO₄²⁻.
3. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Charge on the left is 2+ and on the right is 2+.
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Problem 16 Harder balancing
1. For 2C₄H₁₀: 8CO₂ and 10H₂O, so oxygen on the right is 16 + 10 = 26 and 13O₂ are needed.2. Take 3CO so that 3 carbons give 3CO₂, and oxygen 3 + 3 = 6 on the left and 6 on the right; Fe needs 2.
3. Calcium needs 3Ca(OH)₂ and phosphorus needs 2H₃PO₄. Hydrogen is 6 + 6 = 12, so 6H₂O (oxygen 6 + 8 = 14 and 8 + 6 = 14).
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O; Fe₂O₃ + 3CO → 2Fe + 3CO₂; 3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O
Problem 17 Three precipitations
1. Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s); spectators Na⁺ and Cl⁻; charge 0 = 0.2. Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s); spectators Na⁺ and NO₃⁻; charge 2+ + 2(1−) = 0 = 0.
3. Reaction 1 has spectators Na⁺ and Cl⁻, and reaction 2 has spectators Na⁺ and NO₃⁻. Both ionic equations have zero charge on each side.
Ba²⁺ + SO₄²⁻ → BaSO₄(s); Mg²⁺ + 2OH⁻ → Mg(OH)₂(s)
Problem 18 Deduce the ion charge
1. Total negative charge is 3 × 2− = 6−, so 2 Fe ions carry 6+: each is 3+.2. One O²⁻ means Fe is 2+.
3. 2 × 3− = 6−, so 3 M ions carry 6+: each is 2+.
4. 2 × 3+ = 6+ and 3 × 2− = 6−: Fe₂(CO₃)₃.
Fe³⁺, Fe²⁺, M²⁺ and Fe₂(CO₃)₃.
Problem 19 A displacement reaction
1. Al, CuCl₂, AlCl₃, Cu. Chlorine needs 6 on each side: 2Al(s) + 3CuCl₂(aq) → 2AlCl₃(aq) + 3Cu(s).2. Chloride is the spectator: 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s).
3. Left: 3 × 2+ = 6+. Right: 2 × 3+ = 6+.
2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)
Problem 20 Heating limestone
1. CaCO₃(s) → CaO(s) + CO₂(g).2. 10.0 − 5.6 = 4.4 g.
3. 4.4 ÷ 10.0 × 100 = 44 %.
4. The carbon dioxide gas escapes from the open crucible, so it is no longer weighed, but the total mass of the products is still 10.0 g.
4.4 g of CO₂, which is 44 % of the original mass.
