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 (NH_{3}), we can write a **balanced** chemical equation for it as:

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:

**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:

From the calculation, the smallest mole-to-coefficient ratio is the limiting reactant. Therefore, we can all agree that H_{2} is the limiting reactant. It is the limiting reactant because it is the chemical that is used up first in the reaction, while N_{2} is in excess.

#### If **N**_{2} is in excess, how much of it remains in the product mixture after all the limiting reactant (H_{2}) is used up?

_{2}

To figure this out, we must first determine how much of the excess reactant **(N _{2})** will react with the limiting reactant

**(H**. To do this, we will use the balanced chemical equation to write mole- mole ratios between N

_{2})_{2}and H

_{2}, and then, we will pick the right ratio to use. Here are the ratios:

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:

As we can see, **4 mol** of **N**_{2} will react with **all** the **12 mo**l of **H**_{2} in the starting mixture to form NH_{3}. 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 N_{2} 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 (NH_{3}) 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**_{3 }is 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 H_{2}) is completely used up in the reaction?

If we are both correct, then, we can clearly say that it is particle model **b**. If it’s particle model b, then, the next question is, how much is 8 mol NH_{3} in grams. To determine this, we will multiply the 8 mol NH_{3 }by its molar mass, 17.03 g/mol. Here is a table summarizing the calculation:

Therefore, 136.24 g NH_{3} is the maximum amount of NH_{3 }we can possibly make from 12 mol H_{2}. This maximum amount is usually called the theoretical yield.