Limiting Reagents

 This is part of Year 11 HSC Chemistry course under the topic of Mole Concept

HSC Chemistry Syllabus

  • Explore the concept of the mole and relate this to Avogadro’s constant to describe, calculate and manipulate masses, chemical amounts and numbers of particles in: (ACSCH008, ACSCH039)
– Moles of elements and compounds `n = \frac{m}{MM}` (n = chemical amount in moles, m = mass in grams, MM = molar mass in g mol-1
– Limiting reagent reactions

      Limiting Reagents

      A reagent is termed a limiting reagent when it restricts the quantity of product that can be generated. If a reagent is limiting, the reaction halts once it is entirely consumed.

      A reactant is labeled as in excess if it remains once the limiting reagent has been fully utilis
      ed. This leftover reactant remains unreacted because there are no counterparts left to react with.


      For instance, irrespective of the number of hot dogs, there can be no more than four hot dog-bun combinations. Analogously, in chemistry, if only a limited amount of one reagent is available for a reaction, the reaction ceases when that reagent is depleted, regardless of the quantity of the other reactants.

      How can we identify the limiting reagent?

      To discern the limiting reagent in a reaction, we must comprehend the stoichiometric equivalence in reactions. The technique entails comparing the mole ratio of the reactants involved in the reaction.

      Steps to determine the limiting reagent:

      1. Ascertain the balanced equation for the chemical reaction.
      2. Convert all given weights to moles using `n = \frac{m}{MM}`.
      3. Compute the mole ratio from the balanced equation.
      4. Decide which reactant restricts the amount of product that can be produced.


      Consider the respiration process, a prevalent chemical reaction:

      $$ C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) + \text{energy} $$

      What mass of carbon dioxide is produced in a reaction between 25 grams of glucose and 40 grams of oxygen?

      To address this, note that each mole of glucose requires six moles of oxygen to yield six moles of carbon dioxide and six moles of water.


      1. The reaction equation is already provided.
      2. Using the molar mass of glucose (180.06g/mol), there are 0.13888 moles of glucose. For oxygen (32g/mol), there are 1.25 moles.
      3. To completely utilise 1.25 moles of oxygen, 0.208 moles of glucose are needed `\frac{1.25}{6}`
      4. It's evident that the available glucose is insufficient to react with all the oxygen. Hence, glucose is the limiting reagent, and oxygen is in excess.


      Should the question inquire about excess quantity

      $$ \text{Amount of oxygen left} = 1.25 - (0.13888 \times 6) = 0.4172 \text{ moles} $$