Problem #21: Limiting Reagent Concept [Solved]

1. Explain the meaning of “stoichiometric amounts or equivalent amounts of reactants” in a chemical reaction. Illustrate by an example
2. Explain the concept of “limiting reactant”. Illustrate an example.

Solution:

1. The terms “stoichiometric amounts” and “equivalent amounts” of reactants are often heard when dealing with stoichiometry. Those terms express the same concept. A chemical reaction is represented by a chemical equation (balanced). The chemical equation shows the required and appropriate amounts of the reactants that will react completely together without leaving any excess.

In other words, stoichiometric or equivalent amounts of reactants are the amounts of reactants whose mole ratio matches with the reactant mole ratio in a balanced chemical equation of the reaction.

Please, note that the term “equivalent amounts” is not to be mistaken with “equal amounts”!

Examples

Given the following reaction between sulphuric acid and sodium hydroxide solution:

2NaOH(aq) + H₂SO4(aq) → Na₂SO₄(aq) + 2H₂O(l)

This chemical equation shows us that 2 moles or 80 g of NaOH and 1 mole or 98 g of H₂SO₄ are equivalent/stoichiometric amounts, and they are not equal amounts.

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

In this chemical equation, we see that 1 mole or 40 g of NaOH and 1 mole or 36.5 g of HCl are equivalent/stoichiometric amounts.

2. The concept of a limiting reactant seems to be another concept that causes problems to the students.

Let’s consider the example of the reaction between methane and oxygen to produce carbon dioxide and water:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

1 mole of CH₄ + 2 moles of O₂ → 1 mole of CO₂ + 2 moles of H₂O

In this reaction, the balanced chemical equation shows that the mole ratio of the reactants:

CH₄: O₂ = 1:2

This means that if we react:

16 g of CH₄ and 64 g of O₂, those amounts are stoichiometric quantities because their mole ration, 16g/16 g mol^-: 64 g/32 g mol^- = 1:2, is the same as the mole ratio in the balanced chemical equation. This reaction will produce 44g of CO₂ and 36 g of H₂O.

8 g of CH₄ and 32 g of O₂ are also stoichiometric because 0.5 mole:1 mole = 1:2 mole ratio. This will produce 22 g of CO₂ and 18 g of H₂O.

You notice that although in the two previous chemical equations the quantities change, they remain stoichiometric except that they will produce different quantities of the products. In both cases, the reaction will not leave excess of any reactant.

But if we react:

• 16 g of CH₄ and 32 g of O₂, those quantities are not stoichiometric because their mole ratio 1:1, differs from the mole ratio in the balanced chemical equation, 1:2.

The concept of the “limiting reactant” shows up when the quantities of the reactants are not stoichiometric. The previous example shows that O₂ is the limiting reactant.

Let’s compare this with an example in real life to explain this concept.

A classroom has 25 students; 10 boys and 15 girls. The teacher would like to form groups, each group made of 1 boy and 1 girl. What’s the number of groups that can be formed? 10

The small number of boys limits the number of boy-girl groups that can be formed. Therefore, the boys are a limiting factor in the formation of the boy-girl groups. The number of boy-girl groups is limited to the number of boys, and the remaining 5 girls are in excess.

Compared to the above example, the limiting reactant is the reactant that is completely used up in a reaction, while some quantities of the other reactant(s) still remain. Consequently, it’s the limiting reactant that determines when the reaction stops, and the maximum amount of products that the reaction can produce.

Example: The combustion of CH₄

If in that reaction we have 16 g of CH₄ and 32 g of O₂.

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

1 mole of CH₄ + 2 moles of O₂ (1:2 mole ratio)

But according to the data given, there is 1 mole of CH₄ and 1 mole of O₂, i.e. 1:1 mole ratio!

This shows that the amount of O₂ is lower than its stoichiometric amount, and therefore O₂ is the limiting reactant and CH₄ is in excess. The amount of O₂ will limit the production to 1/2 mole (22 g) of CO₂ and 1 mole (18 g) of H₂O, with an excess of 1/2 mole (8 g) of CH₄.

On the other hand, if we react 16 g (1 mole) of CH₄ with 70 g (2.19 moles) of O₂, then CH₄ is the limiting reactant and oxygen is in excess; 1 mole (44 g) of CO₂ and 2 moles (36 g) of H₂O will be produced, leaving 0.19 mole (6.08 g) of O₂ in excess.

N.B: If you are given a chemical reaction involving two reactants and you are told that one of the reactants is in excess, you must understand that the other reactant is the limiting reactant which will determine the maximum quantity of product(s) expected.

For example, if you burn 8 g (0.5 mole) of methane in excess of oxygen, the methane is the limiting reactant and will limit the production to 0.5 mole (22 g) of carbon dioxide and 1 mole (18 g) of water.