Determining which reactant runs out first

Video tutorial

A chemical equation describes what happens when a chemical reaction takes place. It uses chemical formulae to tell us what happens when the reactants (left) react to form the products.

When glucose is burnt in the open air, there is enough oxygen in the air to ensure that all of the glucose reacts. The oxygen is said to be in excess meaning that there is more than enough of it. The reaction stops when the glucose runs out. When this happens, there is still oxygen left. Glucose is the limiting reactant as the amount of water and carbon dioxide is controlled by how much glucose there is to start with.

If we mix the exact number of moles of each reactant that is needed, they are said to be stoichiometric amounts.

In the glucose example, it was easy to spot that glucose is the limiting reactant since the amount of oxygen in the air is effectively unlimited. In other cases, it may not be so obvious. If there is still some reactant left at the end of a reaction then it is present in excess. In many reactions, such as when there are many reactants or they are colourless gases or solutions, determining the limiting reagent is much more difficult. We then need to work out the number of moles of each reactant and write down the balanced chemical equation:

The limiting reactant is the one that produces the least amount of product.

Example from cooking

Example from cooking:

Consider the following recipe:

1 slice of bread + 2 slices of cheese → 1 cheese toastie

Imagine we have 2 slices of bread and 2 slices of cheese.

• From the recipe, 2 slices of cheese requires 1 slice of bread. There is sufficient bread for this.
• From the recipe, 2 slices of bread requires 4 slice of cheese. There is insufficient cheese for this.
• Hence, cheese is the limiting ingredient/reagent and bread is in excess.
• The maximum meal/yield we can get is limited by the amount of cheese available. As 2 slices of cheese are available, we can make a maximum of 1 cheese toastie. This will use up all of the cheese and leave us with 1 slice of unused/unreacted bread.

Example 1

Example 1:

Consider the following reaction:

C(s) + 2H₂(g) → CH₄(g)

Imagine we have 2 mol of C(s) and 2 mol of H₂(g).

• From the reaction, 2 mol of H₂(g) requires 1 mol of C(s). There is sufficient C(s) for this.
• From the reaction, 2 mol of C(s) requires 4 mol of H₂(g). There is insufficient H₂(g) for this.
• Hence, H₂(g) is the limiting reagent and C(s) is in excees.
• The maximum yield we can get is limited by the amount of H₂(g) available. As 2 mol of H₂(g) are available, we can make 1 mol of CH₄(g). This will use up all of the H₂ and leave us with 1 mol of unreacted C(s).

Example 2

Example 2:

Consider the following reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Imagine we have 2 mol of N₂ and 8 mol of H₂(g).

• From the reaction, 3 mol of H₂(g) requires 1 mol of N₂(g) so 8 mol of H₂(g) requires 8/3 mol of N₂(g). There is insufficient N₂ for this as there is only 2 mol available.
• From the reaction, 1 mol of N₂(g) requires 3 mol of H₂(g) so 2 mol of N₂(g) requires 6 mol of H₂. There is sufficient H₂(g) for this.
• Hence, N₂(g) is the limiting reagent and H₂ is in excees.
• The maximum yield we can get is limited by the amount of N₂(g) available. As 2 mol of N₂(g) is available, we can make 4 mol of NH₃. This will use up all of the N₂(g) and 6 mol of H₂(g), leaving us with 2 mol of unreacted H₂(g).

When working out limiting reagents, moles rather than masses must be used.

Example from cooking

Example from cooking:

Consider the following recipe:

1 slice of bread + 2 slices of cheese → 1 cheese toastie

Imagine we have 80 g of bread and 100 g of chesse, both cut into slides. A slice of bread has a mass of 40 g and a slice of cheese has a mass of 50 g. Clearly, simply using the masses and the recipe would be foolish: we would first convert the masses into numbers of slices and proceed as before.

Example 3

Example 3:

Consider the following reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Imagine we have 100. g of N₂ and 24 g of H₂(g).

• The first step is to convert the masses into moles. N₂ has a molar mass of (2 × 14.01 g mol^-1) = 28.02 g mol^-1:

  number of moles of N₂(g) = mass / molar mass = (100. g) / (28.02 g mol^-1) = 3.57 mol.

  H₂ has a molar mass of (2 × 1.008 g mol^-1) = 2.016 g mol^-1:

  number of moles H₂(g) = mass / molar mass = (24 g) / (2.016 g mol^-1) = 11.9 mol.

• From the reaction, 3 mol of H₂(g) requires 1 mol of N₂(g) so 11.9 mol of H₂(g) requires 11.9/3 mol = 3.97 mol of N₂(g). There is insufficient N₂ for this as there is only 3.57 mol available.
• From the reaction, 1 mol of N₂(g) requires 3 mol of H₂(g) so 3.57 mol of N₂(g) requires 3 × 3.57 mol = 10.7 mol of H₂. There is sufficient H₂(g) for this.
• Hence, N₂(g) is the limiting reagent and H₂ is in excees.
• The maximum yield we can get is limited by the amount of N₂(g) available. As 3.57 mol of N₂(g) is available, we can make 2 × 3.57 mol = 7.14 mol of NH₃. This will use up all of the N₂(g) and 10.7 mol of H₂(g), leaving us with (11.9 - 10.7) mol = 1.2 mol of unreacted H₂(g).

Example 4

Example 4:

From the ideal gas law, PV = nRT, the number of moles of a gas is proportional to the pressure and the volume. For reactions involving gases, we can therefore work with partial pressures or volumes of gases without needing to convert these to moles. Consider the following reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Imagine we have 2.0 L of N₂ and 8.0 L of H₂(g).

• Because the volume of a gas is proportional to the number of moles present, we can read the chemical equation in terms of volumes instead of moles.
• From the reaction, 3.0 L of H₂(g) requires 1.0 L of N₂(g) so 8.0 L of H₂(g) requires 8.0/3.0 L of N₂(g). There is insufficient N₂ for this as there is only 2.0 L available.
• From the reaction, 1.0 L of N₂(g) requires 3.0 L of H₂(g) so 2.0 L of N₂(g) requires 6.0 L of H₂. There is sufficient H₂(g) for this.
• Hence, N₂(g) is the limiting reagent and H₂ is in excees.
• The maximum yield we can get is limited by the amount of N₂(g) available. As 2.0 L of N₂(g) is available, we can make 4.0 L of NH₃. This will use up all of the N₂(g) and 6.0 of H₂(g), leaving us with 2.0 L of unreacted H₂(g).

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