Imagine that you worked at a cycling shop. On a given day you receive a delivery with 20 wheels and 20 frames. Assuming all other components are on hand, how many bicycles can you make?
Obviously you can only make 10, since you need 2 two wheels for every 1 frame. You will be able to make 10 bikes and will have 10 frames left over.
This problem illustrates a common occurrence when running chemical reactions. Some one reactant or building block – the wheel in our case, limits how much product can be made – the bikes in our example. The wheel is a limiting reactant. In this lesson we will determine which reactant is limiting in a given chemical reaction and how much product an be made based on that limitation.
Here is a other example. A box contains 4 hydrogen molecules and 4 oxygen molecules. If the molecules react to form water, which molecules will remain in the box after the reaction is complete?
This is very similar to the bicycle example. We can tell by looking at the molecules, that only 1 oxygen molecule is needed for every 2 hydrogen molecules. This means that we will use up the hydrogens but 2 molecules of oxygen will remain unreacted – they will not find any hydrogens to join with to make water. The 4 hydrogen molecules and the 2 oxygen molecules that react will form 4 water molecules.
Here is a picture of what remains after the reaction is complete.
Here is another way to organize this information without relying on a picture.We write the balanced chemical equation and create three rows below it.
Here is an example. Write the equation and set up the rows. Then..
Step 1: Write down how many molecules of reactants are present in the box.
2H_{2} | + | O_{2} | → | 2H_{2}O | |
have | 4 | 4 | |||
use | |||||
left |
Step 2: Figure out how many molecules can be used. This is the hard step. We can tell by looking at the balanced equation that we need more hydrogen than oxygen molecules. The ratio of H_{2} to O_{2} is 2:1. Since we need more hydrogen than oxygen we write the hydrogen molecules available in the "use" row and follow the 2:1:2 ratio of the balanced equation to write out how many oxygen molecules will react and how many water molecules will be produced.
2H_{2} | + | O_{2} | → | 2H_{2}O | |
have | 4 | 4 | |||
use | 4 | 2 | 4 | ||
left |
Step 3: Figure out how many molecules are left over. This is an easy step. Subtract what is used from what was available on the reactant side, and simply rewrite what was made on the product side.
2H_{2} | + | O_{2} | → | 2H_{2}O | |
have | 4 | 4 | |||
use | 4 | 2 | 4 | ||
left | 0 | 2 | 4 |
Once we know how many molecules can be made we often want to know their mass. In our example the mass of water made can be found by multiplying the water made by the mass per molecule.
$$ 4 \, molecules \times \frac{18 \, u}{ 1 \, molecule} = 72 \, u $$
You may also run into problems which ask for the mass of the remaining material – the excess. Again, we can find the mass of the oxygen in excess by multiplying the number of molecules that are left over by the mass per molecule
$$ 2 \, molecules \times \frac{32 \, u}{ 1 \, molecule} = 64 \, u $$
Here is a other example. A box contains 4 nitrogen molecules and 6 oxygen molecules. If the molecules react to form ammonia, NH_{3}, which molecules will remain in the box after the reaction is complete?
This is slightly more difficult because the ratios are more complex. We can tell by looking at the molecules, that only 1 nitrogen molecule is needed for every 3 hydrogen molecules. This means that we will use up the hydrogens but 2 molecules of nitrogen will remain unreacted – they will not find any hydrogens to join with to make ammonia. The 2 nitrogen molecules and 6 hydrogen molecules that react will form 4 ammonia molecules.
Here is a picture of what remains after the reaction is complete.
The table method will be more practical once larger number of molecules are involved.
Step 1: Write down how many molecules of reactants are present in the box.
N_{2} | + | 3H_{2} | → | 2NH_{3} | |
have | 4 | 6 | |||
use | |||||
left |
Step 2: Figure out how many molecules can be used. This is the hard step. We can tell by looking at the balanced equation that we need 3 times more hydrogen than nitrogen molecules. The ratio of N_{2} to H_{2} is 1:3. Since we need more hydrogen molecules than are available (we would need 4 x 3 = 12 to react all of the nitrogen), the hydrogen is limiting. We write the hydrogen molecules available in the "use" row and follow the 1:3:2 ratio of the balanced equation to write out how many nitrogen molecules will react and how many ammonia molecules will be produced.
N_{2} | + | 3H_{2} | → | 2NH_{3} | |
have | 4 | 6 | |||
use | 2 | 6 | 4 | ||
left |
Step 3: Figure out how many molecules are left over. This is an easy step. Subtract what is used from what was available on the reactant side, and simply rewrite what was made on the product side.
N_{2} | + | 3H_{2} | → | 2NH_{3} | |
have | 4 | 6 | |||
use | 2 | 6 | 4 | ||
left | 2 | 0 | 4 |
Once we know how many molecules can be made it is easy to find their mass. In our example the mass of ammonia made can be found by multiplying the number of ammonia molecules made by the mass per molecule.
$$ 4 \, molecules \times \frac{17 \, u}{ 1 \, molecule} = 68 \, u $$
In the same way we can find the mass of the nitrogen in excess by multiplying the number of molecules that are left over by the mass per molecule.
$$ 2 \, molecules \times \frac{28 \, u}{ 1 \, molecule} = 56 \, u $$