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Limiting & Excess Reactants (DP IB Chemistry): Revision Note

Philippa Platt

Written by: Philippa Platt

Reviewed by: Richard Boole

Updated on

DP Chemistry: HLDP Chemistry: SL

Limiting & Excess Reactants

Excess & limiting reactants

  • Sometimes, there is an excess of one or more of the reactants (excess reactant)

  • The reactant which is not in excess is called the limiting reactant

  • To determine which reactant is limiting:

    • Calculate the number of moles of each reactant

    • This calculation should consider the molar ratio of reactants

Worked Example

What is limiting when 10 mol of carbon are reacted with 3 mol of hydrogen?

C + 2H2 → CH4

Answer:

  • The ratio of C : H2 is 1:2

  • This means that:

    • 10 mol of C requires 20 mol of H2 to fully react

      • There are only 3 moles of H2 available, so H2 is the limiting reactant

    • 3 mol of H2 requires 1.5 mol of C to fully react

      • There are 10 moles of C available

      • So, C is in excess and H2 is the limiting reactant

Examiner Tips and Tricks

To find the limiting reactant quickly, divide the number of moles by the coefficient from the balanced equation:

  • The reactant with the smallest result is limiting

  • The other is in excess

Worked Example

9.2 g of sodium metal is reacted with 8.0 g of sulfur to produce sodium sulfide, Na2S.Which reactant is in excess and which is limiting?

Answer:

Step 1: Calculate the moles of each reactant:

number of moles (Na) = fraction numerator 9.2 space straight g over denominator 22.99 space straight g space mol to the power of negative 1 end exponent end fraction = 0.40 mol

number of moles (S) = fraction numerator 8.0 space straight g over denominator 32.07 space straight g space mol to the power of negative 1 end exponent end fraction = 0.25 mol

Step 2: Write the balanced equation and determine the coefficients

2Na + S → Na2S

Step 3: Divide the moles by the coefficient and determine the limiting reagent

fraction numerator 0.40 space mol space Na over denominator 2 end fraction = 0.20 - lowest

fraction numerator 0.25 space mol space straight S over denominator 1 end fraction = 0.25

Therefore:

  • Sodium is limiting

  • Sulfur is in excess

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Author
Reviewer
Philippa Platt

Author: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

Richard Boole

Reviewer: Richard Boole

Expertise: Chemistry Content Creator

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.

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