AQA GCSE Combined Science (8464), Higher tier · Chemistry › Quantitative chemistry › Use of amount of substance in relation to masses of pure
Practise Limiting reactants (HT only). 15 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.
When two substances react, one is often added in excess so that all of the other is used up. The reactant that is completely used up is the limiting reactant: it decides how much product forms. This subtopic is Higher tier only; you need to identify the limiting reactant from masses or moles and explain its effect on the amount of product.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
5
Define limiting reactant and excessThe limiting reactant is completely used up; the reactant in excess is partly left over.
6
Explain why a reactant is used in excessTo make sure that all of the other reactant is used up.
6
Predict how the limiting reactant affects productThe amount of product is proportional to the amount of limiting reactant; extra excess reactant makes no more product.
7
Identify the limiting reactant using molesDivide the moles of each reactant by its balancing number; the smallest value is limiting.
8
Calculate product mass from the limiting reactantUse only the moles of the limiting reactant in the reacting-mass method.
9
Calculate the mass of excess reactant leftMoles left over = moles at the start − moles that reacted; then convert to a mass.
Notes
Limiting and excess reactants
In a reaction with two reactants, it is common to use an excess of one of them so that all of the other reactant is used up.
The reactant that is completely used up is the limiting reactant, because it limits the amount of product that can form.
The reaction stops when the limiting reactant runs out. Some of the excess reactant is left over.
Doubling the amount of limiting reactant doubles the amount of product (if the other reactant is still in excess). Adding more of the excess reactant makes no more product.
In a practical, if excess magnesium is added to acid, fizzing stops while some magnesium is still left: the acid was the limiting reactant.
Finding the limiting reactant
Step 1: moles of each reactant = mass ÷ Mr.
Step 2: divide each by its balancing number in the equation.
Step 3: the reactant with the smallest answer is the limiting reactant.
Worked example: 4.8 g of Mg is added to 0.30 mol of HCl. Mg + 2HCl → MgCl2 + H2. Mg: 4.8 ÷ 24 = 0.20 mol, and 0.20 ÷ 1 = 0.20. HCl: 0.30 ÷ 2 = 0.15. HCl gives the smaller value, so HCl is limiting and magnesium is in excess.
Working out the product and the leftover
Always use the limiting reactant to calculate the amount of product.
From the example: 0.30 mol of HCl makes 0.15 mol of H2 (ratio 2 : 1), which is 0.15 × 2 = 0.30 g of hydrogen.
Magnesium used = 0.15 mol, so 0.20 − 0.15 = 0.05 mol of Mg is left over. That is 0.05 × 24 = 1.2 g. grade 9+
Cheatsheet
Limiting reactant: completely used up; it limits the amount of product
Excess reactant: some is left over at the end
To find the limiting reactant: moles ÷ balancing number; the smallest value is limiting
Amount of product is proportional to the amount of limiting reactant
Calculate the product from the limiting reactant only
Excess left over = moles at start − moles reacted
How to answer each type of question
Explain why a reactant is used in excess
1 to 2 marks6
Say that it makes sure all of the other reactant is used up.
If it fits the context, say how the leftover excess is removed (e.g. by filtering).
Example. To make copper sulfate solution, a student adds copper oxide to warm sulfuric acid until no more dissolves, so that the copper oxide is in excess. Explain why the copper oxide is added in excess. [2 marks]
Show the model answer
So that all of the sulfuric acid reacts / is used up (1). The unreacted copper oxide can then be removed by filtering, leaving only copper sulfate solution (1).
Explain the effect of changing the amount of a reactant
2 marks6
Identify which reactant is limiting.
Say that the amount of product is proportional to the amount of limiting reactant.
Example. A student reacts excess marble chips with 50 cm3 of hydrochloric acid and collects the carbon dioxide. She repeats the experiment with 100 cm3 of the same acid. The marble chips are still in excess. Explain the effect on the total volume of carbon dioxide produced. [2 marks]
Show the model answer
The acid is the limiting reactant (1). Doubling the amount of acid doubles the amount of carbon dioxide, so the volume doubles (1).
Identify the limiting reactant
3 marks7
Work out the moles of each reactant.
Compare them using the balancing numbers in the equation.
State clearly which reactant is limiting and which is in excess.
Example. Zinc reacts with copper sulfate solution. Zn + CuSO4 → ZnSO4 + Cu A student added 1.30 g of zinc to a solution containing 0.0250 mol of copper sulfate. Show which reactant is the limiting reactant. Ar: Zn = 65 [3 marks]
Show the model answer
moles of Zn = 1.30 ÷ 65 = 0.0200 mol (1) the ratio is 1 : 1, and 0.0200 mol is less than 0.0250 mol (1) so zinc is the limiting reactant (copper sulfate is in excess) (1)
Calculate the mass of product from the limiting reactant
4 marks8
Work out the moles of both reactants.
Use the equation to decide which is limiting.
Use the limiting reactant's moles and the ratio to find the moles of product.
Convert to a mass.
Example. Hydrogen reacts with chlorine. H2 + Cl2 → 2HCl 1.0 g of hydrogen is mixed with 14.2 g of chlorine. Calculate the maximum mass of hydrogen chloride that can form. Ar: H = 1; Cl = 35.5 [4 marks]
Show the model answer
moles of H2 = 1.0 ÷ 2 = 0.50; moles of Cl2 = 14.2 ÷ 71 = 0.20 (1) the ratio is 1 : 1, so chlorine is limiting (1) moles of HCl = 2 × 0.20 = 0.40 mol (1) mass of HCl = 0.40 × 36.5 = 14.6 g (1)
Shortcuts and memory tricks
Divide by the balancing number: moles ÷ balancing number, and the smallest value is limiting.
Think of sandwiches: 10 slices of bread and 3 slices of cheese make only 3 sandwiches (2 bread + 1 cheese each). The cheese is limiting and the bread is in excess.
In a practical, the reactant you can still see at the end (solid left over, colour remaining) is the one in excess.
More limiting reactant = more product. More excess reactant = no more product.
Where marks are lost
Choosing the reactant with the smaller mass as limiting. You must compare moles, using the ratio.
Forgetting the balancing numbers when comparing moles.
Calculating the amount of product from the excess reactant.
Saying the excess reactant 'does not react'. Some of it reacts; only the leftover part is in excess.
Exam technique
Show the moles of both reactants and state clearly which one is limiting.
Use the phrases 'completely used up' and 'in excess' in explanations.
When a question says one reactant is in excess, the other one is limiting: base your calculation on it.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
Hydrogen reacts with chlorine: H2 + Cl2 → 2HCl In a reactor, 3 mol of hydrogen was mixed with 2 mol of chlorine and the mixture reacted. Which reactant is the limiting reactant?
chlorine
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
In many reactions, one of the reactants is used in excess.
(a) Complete the sentence. Tick (✓) one box. The reactant that is completely used up in a reaction is called the ...[1]
excess reactant.
limiting reactant.
catalyst.
product.
(b) Give one reason why one reactant is often used in excess.[1]
(c) Hydrogen reacts with chlorine: H2 + Cl2 → 2HCl In a reactor, 3 mol of hydrogen was mixed with 2 mol of chlorine and the mixture reacted. Which reactant is the limiting reactant?[1]
(d) How many moles of hydrogen chloride are produced from 3 mol of hydrogen and 2 mol of chlorine?[1]
Show the answer and mark scheme
(a)Answer: limiting reactant.
(b)
to make sure that all of the other reactant is used up / reacts
(c)Answer: chlorine
chlorine / Cl2
(d)Answer: 4 mol
4 (mol)
Question 2Medium6 marks
Zinc reacts with copper(II) sulfate solution: Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) A student added 1.30 g of zinc powder to a blue solution containing 0.025 mol of copper(II) sulfate. Relative atomic masses (Ar): Cu = 63.5, Zn = 65
(a) Calculate the number of moles of zinc added.[1]
(b) Explain which reactant is the limiting reactant.[2]
(c) Calculate the maximum mass of copper that can be produced.[2]
(d) Suggest what the student would see at the end of the reaction that shows copper(II) sulfate was in excess.[1]
Show the answer and mark scheme
(a)Answer: 0.02 mol
1.30 ÷ 65 = 0.02 (mol)
(b)
zinc and copper(II) sulfate react in a 1 : 1 ratio and there are fewer moles of zinc (0.02 mol) than copper(II) sulfate (0.025 mol)
so zinc is the limiting reactant / copper(II) sulfate is in excess
(c)Answer: 1.27 g
moles of Cu = 0.02
0.02 × 63.5 = 1.27 (g)
(d)
the solution is still (pale) blue
Question 3Hard8 marks
Aluminium reacts with chlorine to produce aluminium chloride: 2Al + 3Cl2 → 2AlCl3 A student reacted 5.4 g of aluminium with 14.2 g of chlorine. Relative atomic masses (Ar): Al = 27, Cl = 35.5
(a) Show that chlorine is the limiting reactant.[3]
(b) Calculate the maximum mass of aluminium chloride that can be produced. Give your answer to 3 significant figures.[3]
(c) Calculate the mass of aluminium left over at the end of the reaction.[2]
Show the answer and mark scheme
(a)
moles of Al = 5.4 ÷ 27 = 0.2
moles of Cl2 = 14.2 ÷ 71 = 0.2
0.2 mol Al would need 0.3 mol Cl2 (ratio 2 : 3) but there is only 0.2 mol Cl2, so chlorine is limiting