Higher tier

# Limiting reactions

Magnesium ribbon reacts with hydrochloric acid according to the equations below:

##### Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
Most of the time when carrying out a chemical reaction it is common to use an excess of one reactant (this simply means to use more than you actually need to react), this will ensure that the other reactant is completely used up. In the reaction of magnesium with hydrochloric acid to ensure that all the magnesium ribbon reacts use an excess of the hydrochloric acid, that is use much more hydrochloric acid than you actually need to ensure that all the magnesium reacts and none is left, the diagram below explains this:

## The limiting factor in a chemical reaction

The reactant which is not in excess will be the one that will determine the maximum amount of product that can be obtained- it is called the limiting factor or limiting reactant, for example magnesium ribbon burns in air to form magnesium oxide as shown in the image opposite, here the air is clearly in excess and the limiting factor, that is the factor that will determine how much of the product is formed is obviously the amount of magnesium ribbon. It is the mass of magnesium that will determine the amount of magnesium oxide produced.

To calculate the mass of magnesium oxide formed when a given mass of magnesium is burned simply use the method shown here. In the exam just make sure you identify which of the reactants is in excess as it is this one that will determine the maximum mass of product or yield of reaction.

Example 1: Calculate the maximum mass of magnesium oxide obtained by burning 10g of magnesium ribbon in air.
Step 1- You need to write a balanced symbolic equation for the reaction, more than likely you will be given one in the exam!

##### 2Mg(s) + O2(g) → 2MgO(s)

Step 2- Calculate the relative formula masses of the magnesium oxide, we don't need to bother with oxygen as it is in excess. It is the amount of magnesium that will determine the mass of the product obtained, it is the limiting factor here.
Ar of Mg=24 Ar of O=16
##### 2Mg(s) → 2MgO(s)

Remember the Ar or Mr expressed in grams will give the mass of 1 mole of the substance. So here 2 moles or 48g of magnesium will produce 2 moles or 80g of magnesium oxide. Or simply 1 mole of magnesium will produce 1 mole of magnesium oxide; the ratio of reactant to product is 1:1.

However in this example we are only burning 10g of magnesium, not 48g. So simply scale down. I would suggest to begin with you calculate the mass of magnesium oxide produced by burning 1g of magnesium:

##### 24g → 40g
So divide both sides of the equation by 24, since any number divided by itself is 1, this will give the mass of magnesium oxide produced by burning 1g of magnesium.
##### 1g → 1.67g
So burning 1g of magnesium produces 1.67g of magnesium oxide, in this example the question asked how much magnesium oxide is produced by burning 10g of magnesium, so simply multiply 1.67 x 10= 16.7g of magnesium oxide produced.

### Identifying the limiting factor

To identify the limiting factor in a reaction consider the reaction of nitrogen gas (N2) with hydrogen gas (H2) to make ammonia (NH3); word and symbolic equations for this reaction are shown below:

##### N2(g) + 3H2(g) ⇌ 2NH3(g)

Now if a student mixes 10 moles of nitrogen gas with 15 moles of hydrogen gas then which reactant is the limiting factor in this reaction and how much ammonia gas is produced?

Well from the balanced symbolic equation above we can see that 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas, so in order for both gases to be completely consumed and turned into ammonia they need to be mixed in the ratio of 1:3. Now to work out which reactant is the limiting factor and which one is in excess simply calculate the moles of product that would be formed from each reactant; as shown below:
• For nitrogen gas (N2 ): 10 moles of nitrogen would need 30 moles of hydrogen to react completely with it since they react in the ratio of 1:3.
• and for hydrogen gas we have:
• 15 moles of hydrogen would need 5 moles of nitrogen to react completely with and 10 moles are available.

So the 10 moles of nitrogen gas required 30 moles of hydrogen gas to react completely with but only 15 moles are available, so the nitrogen gas is in excess and the hydrogen gas will be the limiting reactant; that is the reactant that will limit how much ammonia can be made.

To calculate the amount of ammonia that will be produced we simply need to look at the balanced symbolic equation again. From this we can see that 3 moles of hydrogen gas will produce 2 moles of ammonia, or by dividing by 3 we can simply say that 1 mole of hydrogen will form 2/3 or 0.67 moles of ammonia. So the 15 moles of hydrogen that the student added will form 10 moles of ammonia gas (15 x 0.67= 10 moles)

#### Example 2

Aluminium will react very violently with chlorine gas to form a white solid called aluminium chloride. A balanced symbolic equation for this reaction is shown below:

##### 2Al(s) + 3Cl2(g) → 2AlCl3(s)

From this symbolic equation you can see that 2 moles of aluminium react with 3 moles of chlorine gas to form 2 moles of aluminium chloride. Let's simplify the moles ratios in the symbolic equation down as much as possible and we can then see that 1 mole of aluminium metal requires 1.5 moles of chlorine gas to react with, when these two reactants are mixed in the ratio of 1 mole: 1.5 mole they will both consume each other completely and form 1 mole of the product; aluminium chloride.
##### 1 mole 1.5 moles

So let's imagine that a student mixes 4 moles of aluminium metal with 5 moles, of chlorine gas; then what is the limiting factor here and how much aluminium chloride will be formed?
##### 4 moles 5 moles
Now we already know that the ideal ratio for aluminium and chlorine to be mixed together at is 1 mole: 1.5 moles, so if we have 4 moles of aluminium metal then it will require 4 x 1.5 moles of chlorine gas, that is 6 moles of chlorine gas to react with. However the student has only mixed in 5 moles of chlorine gas. This means that some of the aluminium metal will be left unreacted since there is not enough chlorine to react with it all; that is chlorine is the limiting factor in this reaction; there is simply not enough chlorine gas present to react with all the aluminium and this will obviously limit the amount of aluminium chloride that can be made. To work out how much aluminium chloride will be formed we can go back to the symbolic equation for the reaction but this time simplify it down by removing the aluminium metal from it, since it is in excess:
##### 2Al(s) + 3Cl2(g) → 2AlCl3(s)

Now from this equation we can see that 3 moles of chlorine gas will form 2 moles of aluminium chloride:
##### 3 moles of chlorine → 2 moles of aluminium chloride
Now we can simplify this and say 1 mole of chlorine will form 1.5 moles of aluminium chloride:
##### 3 moles of chlorine → 2 moles of aluminium chloride
or simply by dividing both sides by 3:
##### 1 mole of chlorine → 0.67 moles of aluminium chloride
So if the student added 5 moles of chlorine gas then it will produce 5 x 0.67 moles, that is 3.33 moles of aluminium chloride:

### Key points

• If a reactant is said to be in excess it simply means there is more than enough of it to react with any other reactant or reactants present. The main reason a reactant maybe in excess is to ensure that the other reactant is completely used up.
• The limiting reactant is the one that is all used up or is consumed in a chemical reaction. It is the amount of the limiting reactant present that will determine how much product can be made.