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.**

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.

Consider the following reaction:
C(s) + 2H_{2}(g) → CH_{4}(g)

Imagine we have 2 mol of C(s) and 2 mol of H_{2}(g).
- From the reaction, 2 mol of H
_{2}(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 H
_{2}(g). There is insufficient H_{2}(g) for this.
- Hence, H
_{2}(g) is the limiting reagent and C(s) is in excees.
- The maximum yield we can get is limited by the amount of H
_{2}(g) available. As 2 mol of H_{2}(g) are available, we can make 1 mol of CH_{4}(g). This will use up all of the H_{2} and leave us with 1 mol of unreacted C(s).

Consider the following reaction:
N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)

Imagine we have 2 mol of N_{2} and 8 mol of H_{2}(g).
- From the reaction, 3 mol of H
_{2}(g) requires 1 mol of N_{2}(g) so 8 mol of H_{2}(g) requires 8/3 mol of N_{2}(g). There is insufficient N_{2} for this as there is only 2 mol available.
- From the reaction, 1 mol of N
_{2}(g) requires 3 mol of H_{2}(g) so 2 mol of N_{2}(g) requires 6 mol of H_{2}. There is sufficient H_{2}(g) for this.
- Hence, N
_{2}(g) is the limiting reagent and H_{2} is in excees.
- The maximum yield we can get is limited by the amount of N
_{2}(g) available. As 2 mol of N_{2}(g) is available, we can make 4 mol of NH_{3}. This will use up all of the N_{2}(g) and 6 mol of H_{2}(g), leaving us with 2 mol of unreacted H_{2}(g).

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

Consider the following reaction:
N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)

Imagine we have 100. g of N_{2} and 24 g of H_{2}(g).
- The first step is to convert the masses into moles. N
_{2} has a molar mass of (2 × 14.01 g mol^{-1}) = 28.02 g mol^{-1}:
number of moles of N_{2}(g) = mass / molar mass = (100. g) / (28.02 g mol^{-1}) = 3.57 mol.

H_{2} has a molar mass of (2 × 1.008 g mol^{-1}) = 2.016 g mol^{-1}:
number of moles H_{2}(g) = mass / molar mass = (24 g) / (2.016 g mol^{-1}) = 11.9 mol.

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

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:
N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)

Imagine we have 2.0 L of N_{2} and 8.0 L of H_{2}(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 H
_{2}(g) requires 1.0 L of N_{2}(g) so 8.0 L of H_{2}(g) requires 8.0/3.0 L of N_{2}(g). There is insufficient N_{2} for this as there is only 2.0 L available.
- From the reaction, 1.0 L of N
_{2}(g) requires 3.0 L of H_{2}(g) so 2.0 L of N_{2}(g) requires 6.0 L of H_{2}. There is sufficient H_{2}(g) for this.
- Hence, N
_{2}(g) is the limiting reagent and H_{2} is in excees.
- The maximum yield we can get is limited by the amount of N
_{2}(g) available. As 2.0 L of N_{2}(g) is available, we can make 4.0 L of NH_{3}. This will use up all of the N_{2}(g) and 6.0 of H_{2}(g), leaving us with 2.0 L of unreacted H_{2}(g).

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