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1.2.6Building Blocks for Understanding

Limiting Reactants (HT Only) — AQA GCSE Combined Science Synergy 8465 - Physical Sciences Revision Notes & Topical Questions

5 min read 8 parts Lesson 12 of 14

📚 AQA GCSE Combined Science Synergy 8465 - Physical Sciences Hub / Topical Revision
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Part 1 of 6Limiting and excess (HT only)

Limiting reactant
  • Limiting reactant The reactant that is completely used up first, so the reaction stops and no more product can form.
  • It is the reactant that limits how much product is made.
Excess reactant
  • Excess reactant Any reactant left over once the reaction has finished.
  • It is present in a larger amount than the equation requires.
Why it matters
  • Why it matters The amount of product depends on the limiting reactant,
    • so industry often uses an excess of a cheap reactant to ensure an expensive one is fully used.
  • This also helps to avoid wasting costly materials.

Part 2 of 6Everyday picture (HT only)

1Each sandwich needs one slice of cheese and two of bread
  • Each sandwich needs one slice of cheese and two of bread Imagine 10 slices of bread and 3 slices of cheese.
2Cheese runs out first
  • Cheese runs out first Only 3 sandwiches can be made, so cheese is limiting even though there seems to be plenty of bread.
3Bread left over
  • Bread left over Six of the ten slices are used and four remain in excess.

Part 3 of 6Finding the limiting reactant (HT only)

1
Write the balanced equation.
2
Convert the mass of each reactant to moles using moles = mass ÷ Mr.
3
Compare the moles with the equation ratio, by dividing each amount by its coefficient.
4
The smaller result belongs to the limiting reactant.
5
Use the limiting reactant's moles and the equation ratio to calculate the product amount.
6
Convert the moles of product to mass if the question asks for it.

Part 4 of 6Worked example (HT only)

StepWorking
Equation2Mg + O2 → 2MgO
Given4.8 g Mg (Ar 24) and 6.4 g O2 (Mr 32)
MolesMg = 4.8 ÷ 24 = 0.20 mol, O2 = 6.4 ÷ 32 = 0.20 mol
Divide by coefficientMg = 0.20 ÷ 2 = 0.10, O2 = 0.20 ÷ 1 = 0.20, so Mg is limiting
Product0.20 mol MgO, mass 0.20 × 40 = 8.0 g
Excess leftO2 used = 0.10 mol, so 0.10 mol (3.2 g) of O2 remains

Part 5 of 6Linked ideas (HT only)

Limiting reactant
Gas volume
  • Gas volume In an acid and metal reaction the volume of hydrogen collected levels off once the limiting reactant is gone.
Graph shape
  • Graph shape A mass or volume against time graph becomes flat when the limiting reactant is used up.
Doubling the limiting reactant
  • Doubling the limiting reactant Doubling the amount of the limiting reactant doubles the final amount of product,
    • but adding more excess reactant changes nothing.
Percentage yield
  • Percentage yield The maximum possible yield is always worked out from the limiting reactant.

Part 6 of 6Exam technique for limiting reactant questions (HT only)

Show the division
  • Show the division Always write the moles of each reactant and then the value after dividing by the coefficient,
    • so the examiner can see why one reactant is limiting.
State your conclusion
  • State your conclusion Write a clear sentence such as 'magnesium is the limiting reactant because it has the smaller value', rather than leaving the answer as numbers alone.
Use the correct substance
  • Use the correct substance All later product calculations must start from the moles of the limiting reactant, never from the excess reactant,
    • which is only partly used.

Key terms

TermMeaning
Limiting reactantThe reactant that is used up first and determines the amount of product.
ExcessMore of a reactant than is needed to react with the limiting reactant.
YieldThe amount of product obtained from a reaction.
Mole ratioThe ratio of moles shown by the coefficients in an equation.
CoefficientThe large number in front of a formula in a balanced equation.

Avoid these mistakes

1
Choosing the reactant with the smaller mass as limiting without converting to moles.
2
Not dividing by the coefficient before comparing the moles.
3
Using the excess reactant to calculate the mass of product.
4
Thinking the reaction continues after the limiting reactant has run out.
5
Forgetting that the leftover excess can be calculated by subtracting the amount used from the amount supplied.

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