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3.2.3Using moles to balance equations

AQA GCSE Chemistry (8462), Higher tier · Quantitative chemistry › Moles and masses

Practise Using moles to balance equations. 11 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Sample questions

Written for this site in the style of AQA exam questions. They are not taken from real past papers.

Question 1Easy4 marks
The balancing numbers in an equation can be found from the numbers of moles of the substances that react and are produced.
number of moles = mass in g ÷ Mr
(a) In a reaction, 0.2 mol of hydrogen, H2, reacted with 0.1 mol of oxygen, O2, to produce 0.2 mol of water, H2O.
Give the simplest whole number ratio of moles of H2 : O2 : H2O.[1]
(b) Use this ratio to write the balanced equation for the reaction.[1]
(c) Calculate the number of moles in 4.4 g of carbon dioxide.
Mr of CO2 = 44[1]
(d) In another reaction, 0.3 mol of sodium reacted with 0.15 mol of chlorine, Cl2, to produce 0.3 mol of sodium chloride, NaCl.
Write the balanced equation for this reaction.[1]
Show the answer and mark scheme
(a) Answer: 2 : 1 : 2
  • 2 : 1 : 2
(b) Answer: 2H2 + O2 → 2H2O
  • 2H2 + O2 → 2H2O
(c) Answer: 0.1 mol
  • 0.1 (mol)
(d) Answer: 2Na + Cl2 → 2NaCl
  • 2Na + Cl2 → 2NaCl
Question 2Medium5 marks
A student burned 2.4 g of magnesium in oxygen. 1.6 g of oxygen reacted and 4.0 g of magnesium oxide was produced.
Relative atomic masses (Ar): O = 16, Mg = 24
(a) Calculate the number of moles of magnesium (Mg), oxygen (O2) and magnesium oxide (MgO) involved in the reaction.[3]
(b) Use your answers to write the balanced equation for the reaction.[2]
Show the answer and mark scheme
(a) Answer: Mg 0.1 mol, O2 0.05 mol, MgO 0.1 mol
  • Mg: 2.4 ÷ 24 = 0.1 (mol)
  • O2: 1.6 ÷ 32 = 0.05 (mol)
  • MgO: 4.0 ÷ 40 = 0.1 (mol)
(b) Answer: 2Mg + O2 → 2MgO
  • ratio 0.1 : 0.05 : 0.1 = 2 : 1 : 2
  • 2Mg + O2 → 2MgO
Question 3Hard6 marks
Iron reacts with chlorine to form an iron chloride.
A student found that 5.60 g of iron reacted completely with 10.65 g of chlorine, Cl2.
Relative atomic masses (Ar): Cl = 35.5, Fe = 56
(a) Calculate the mass of iron chloride produced.[1]
(b) The relative formula mass of the iron chloride is 162.5.
Show that the formula of the iron chloride is FeCl3.[1]
(c) Use the masses to find the balancing numbers in the equation for the reaction.
Write the balanced equation.[4]
Show the answer and mark scheme
(a) Answer: 16.25 g
  • 16.25 (g)
(b)
  • 56 + (3 × 35.5) = 162.5
(c) Answer: 2Fe + 3Cl2 → 2FeCl3
  • moles of Fe = 5.60 ÷ 56 = 0.1
  • moles of Cl2 = 10.65 ÷ 71 = 0.15
  • moles of FeCl3 = 16.25 ÷ 162.5 = 0.1
  • ratio 2 : 3 : 2 so 2Fe + 3Cl2 → 2FeCl3

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