26 Mar
Posted by: sean in: Chemistry Notes & Tips
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Concepts of Limiting & Excess Reactants are so commonly tested in school exams as well as ‘O’ Levels Chemistry examinations. However, many students seemed to have very weak foundation of the essential concepts surrounding it.
Before we even visit any question, it is important to understand the definition of both reactants.
Limiting Reactant:
- The reactant that is completely used up in a chemical reaction
- It determines or limits the amount of products formed in a reaction
Excess Reactant:
- The reactant that are not used up in a chemical reaction
- There is always an excess at the end of the reaction, stays unreacted
Example:
Reaction of hydrogen gas with chlorine according to the equation below:
H2 (g) + Cl2 (g) –> 2HCl (g)
We can see that 1 mole of H2 reacts with 1 mole of Cl2 to form 2 moles of HCl
Scenario A: When 1 mole of H2 added to 1 mole of Cl2
Results: 2 moles of HCl are produced and all H2 & Cl2 are used up.
Conclusion: No Limiting & Excess Reactants.
Scenario B: When 2 moles of H2 added to 1 mole of Cl2
Results: 2 moles of HCl are produced and 1 mole of H2 stays unreacted.
Conclusion: Cl2 is the Limiting Reactant & H2 is the Excess Reactant
Scenario C: When 1 mole of H2 added to 2 moles of Cl2
Results: 2 moles of HCl are produced and 1 mole of Cl2 stays unreacted.
Conclusion: Cl2 is the Excess Reactant & H2 is the Limiting Reactant
Impt Notes:
1. Must always have a balanced chemical equations first
2. Compare ”No. of Moles” between the 2 reactants to determine which is Limiting / Excess
3. If all reactants & products are in gaseous state, you can compare by volume directly (if given)
Let me know if the above explanation is useful to you. I would love to hear from you.
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5 Responses
SimpleChemConcepts | O Level Chemistry: Mole Calculations – Limiting & Excess Reactants
28|Mar|2010 1[...] the previous blogpost, we have discussed on the concepts surrounding Limiting & Excess Reactants, as well as look [...]
ambra
16|Apr|2010 2it has beinb very useful to me
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sean Reply:
May 6th, 2010 at 12:43 am
Thanks for encouragement. Keep learning!
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abrar
25|Apr|2010 3it would have been more useful if u would have solved a tougher question concerning moles rather than the one answered above.
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sean Reply:
May 6th, 2010 at 12:43 am
Check out the archive blogposts on discussions based on Mole Concepts and Mole Calculations.
Rgds,
Sean Chua
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