Identifying The Limiting Reactant In A Drawing Of A Mixture

In chemical reactions, reactants are converted into products. In other words, atoms of reactants in the starting mixture regroup in new ways to form new chemicals called products in the product mixture. Since we can't see atoms with our eyes, one way we can make sense of these atoms reacting is to consider them as particles interacting with one another. Because of this, when hydrogen molecules (H ₂) react with nitrogen molecules (N ₂ ) to form ammonia molecules (NH₃), we can write a balanced chemical equation for it as:

Chemical equation for synthesis of ammonia — Chemical equation

Now, imagine that before the reaction starts, we zoomed in really close and saw molecules of H ₂ and N ₂, which we drew its particle model as:

Hydrogen and nitrogen molecules — Starting reaction mixture

How do we interpret the particle model in terms of moles?

To interpret the particle model in terms of moles, we will count the number of each type of molecule in the model and then attach moles to the number. If we count, you will notice that we have 12 moles of hydrogen (H ₂) molecules and 6 moles of nitrogen (N ₂) molecules.

Now, which of the two chemicals is the limiting reactant and which is in excess?

To determine the limiting reactant, we will divide the moles of each reactant by its corresponding coefficient in the balanced chemical equation:

From the equation, you will notice the coefficient (number in front of chemical symbol in equation) of N ₂ is 1 and the coefficient of H ₂ is 3.

If we divide the moles of each reactant from the particle model by its corresponding coefficient from the balanced equation, we will get:

Finding mole to coefficient ratio by dividing moles of chemical by its corresponding coefficient in balanced equation — Finding limiting reactant

From the calculation, the smallest mole-to-coefficient ratio is the limiting reactant. Therefore, we can all agree that H₂ is the limiting reactant. It is the limiting reactant because it is the chemical that is used up first in the reaction, while N₂ is in excess.

IfN₂ is in excess, how much of it remains in the product mixture after all the limiting reactant (H₂) is used up?

To figure this out, we must first determine how much of the excess reactant(N₂) will react with the limiting reactant(H₂). To do this, we will use the balanced chemical equation to write mole- mole ratios between N₂ and H₂, and then, we will pick the right ratio to use. Here are the ratios:

Mole ratio between N and H — Writing mole-mole ratio

Since we want to figure out the moles of N ₂ that will react with H ₂, we will pick the ratio with N ₂ on top. Therefore, the complete setup for our calculation will appear as:

moles of nitrogen — Calculating moles of N

As we can see, 4 mol of N ₂ will react withall the 12 mol of H ₂ in the starting mixture to form NH₃. So, the next question is, how much of the excess reactant is still in the product mixture after the reaction is complete? To determine this, we will simply subtract 4 mol of N ₂ from the starting moles of N₂ in the starting mixture. If we do, we will get 2 moles; That is: 6 mol N ₂ – 4 mol N ₂ .

The next question is, how many moles of ammonia (NH₃) will be formed after all the limiting reactant is used up?

To determine this, we will convert moles of the limiting reactant (H ₂) to moles of NH ₃. To do this, we will use the balanced chemical equation to write the mole-mole ratio between H ₂ and NH _3, and then, setup a table similar to the one above. Here is the table:

As you can see, 8 mol NH ₃is expected to form after all the 12 mol H ₂ completely reacts with the 4 mol N ₂.

Now, let's test our understanding.

Which of the following particle models depicts the product mixture after all the limiting reactant (12 mol H₂) is completely used up in the reaction?

Represents ammonia and nitrogen molecules — Particle models

If we are both correct, then, we can clearly say that it is particle modelb. If it's particle model b, then, the next question is, how much is 8 mol NH₃ in grams. To determine this, we will multiply the 8 mol NH₃by its molar mass, 17.03 g/mol. Here is a table summarizing the calculation:

Therefore, 136.24 g NH₃ is the maximum amount of NH₃we can possibly make from 12 mol H₂. This maximum amount is usually called the theoretical yield.