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Chapter 4 SABIS Grade 10 Part 7

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Lesson 22: Limiting Reagents and Excess Reagents 🧪🥽🔍


In chemistry, reactions 🔄 don't always use up all the reactants equally. Sometimes, one reactant gets used up first. That reactant is called the Limiting Reagent because its amount limits the amount of product that can be formed. The reactant that is not completely used up is called the Excess Reagent.

Let's get a grasp on this concept with a few basic questions 🤔:

Basic Question 7: Limiting and Excess Reagents

Consider the following balanced chemical equation:

4Fe + 3O2 → 2Fe2O3

If you started the reaction with 5 moles of Fe and 4 moles of O2, which one is the limiting reagent? Which one is in excess? How many moles of Fe2O3 can be formed? How many moles of the excess reagent will be left over at the end of the reaction?

  1. Given: 5 moles Fe and 4 moles O2

  2. For Fe: 5 moles Fe x (2 moles Fe2O3 / 4 moles Fe) = 2.5 moles Fe2O3

  3. For O2: 4 moles O2 x (2 moles Fe2O3 / 3 moles O2) = 2.67 moles Fe2O3

The smaller value will determine the amount of product formed and hence Fe is the limiting reagent here. Using up 5 moles of Fe will consume 3.75 moles of O2 (from 5 moles Fe x 3 moles O2 / 4 moles Fe), leaving 0.25 moles of O2 unused, so O2 is the excess reagent. We can form 2.5 moles of Fe2O3 based on the amount of the limiting reagent.

Let's try another example to strengthen our understanding of this concept:

Additional Example

Consider the following balanced chemical equation:

2H2 + O2 → 2H2O

If you started the reaction with 5 moles of H2 and 3 moles of O2, which one is the limiting reagent? Which one is in excess? How many moles of H2O can be formed? How many moles of the excess reagent will be left over at the end of the reaction?

  1. Given: 5 moles H2 and 3 moles O2

  2. For H2: 5 moles H2 x (2 moles H2O / 2 moles H2) = 5 moles H2O

  3. For O2: 3 moles O2 x (2 moles H2O / 1 mole O2) = 6 moles H2O

The smaller value will determine the amount of product formed and hence H2 is the limiting reagent here. Using up 5 moles of H2 will consume 2.5 moles of O2 (from 5 moles H2 x 1 mole O2 / 2 moles H2), leaving 0.5 moles of O2 unused, so O2 is the excess reagent. We can form 5 moles of H2O based on the amount of the limiting reagent.

Great work! You've got this! 💪😁

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