Limiting reactants and percent yield

Transcript

Content Reviewers:

Rishi Desai, MD, MPH

Limiting reactants limit the amount of products a chemical reaction can produce.

Let's use a simple example.

Let’s say we have 10 people and 5 chairs.

If only one person can sit in a chair, then only five people can sit down at a time while five others stand.

We can write this as a chemical equation:

1 Person + 1 Chair —> 1 Sitting Person

So because it takes one person, and one chair, to produce one sitting person, then we know that with 5 chairs and 10 people we are limited to producing at most 5 sitting people.

So the number of chairs is the limiting reactant, or limiting reagent, in our example.

But sometimes figuring out the limiting reactant is not very intuitive.

And so we will use the chair example to learn how to find the limiting reactant more systematically, by using mole ratios.

If we want to analyze this problem using mole ratios, we start by looking at the coefficients in the balanced chemical equation.

In the equation, the coefficients of “Person” and “Chair” are both one---so the mole ratio between the reactants is one-to-one:

Person/Chair = 1/1

Now that we have the mole ratio between the reactants, we pick either one of the numbers that we are given in our problem---we can pick either 5 chairs, or we can pick 10 people, it does not matter.

Here, suppose we pick 10 people.

We write this number into our mole ratio equation, and then put an “x” in place of the number of chairs

Person/Chair = 1/1 = 10/x

If we solved the equation, we would find that “x” is equal to “10”

This represents how many chairs would be needed to have a perfectly balanced reaction with 10 people.

So, since we know from our problem that we only have 5 chairs, we now know that chairs must be our limiting reactant---we have fewer chairs than we need to react with every person.

Let’s say that instead of picking 10 people as our first number, we instead picked 5 chairs.

In this case, we would write out the mole ratio with an unknown “y” in place of the number of people:

Person/Chair = 1/1 = y/5

We can solve this equation in order to get y equals 5, meaning that we would need exactly 5 people to react with 5 chairs.

However, we know that we actually have 10 people, meaning that the number of people is in excess---we have more people than chairs.

Once again, this would make chairs the limiting reactant in this problem.

We can also set this problem up by drawing a little table.

Once we know our limiting reactant, drawing a table helps us figure out how much product we will make.

The columns of our table correspond to each part of our chemical reaction: here, these are people, chairs, and sitting people.

We label the rows with “I” for “Initial”, “C” for “Change”, and “F” for “Final.”

In the “initial” row, we put down all the values we start with, before the reaction takes place.

So we write down numbers corresponding to 10 people, 5 chairs, and zero Sitting people.

We just determined that chairs were our limiting reactant, and so all of the chairs get used up in the reaction.

So in the “Change” row (what happens during the reaction), we put a “minus five” in the column for chairs.

These chairs reacted with 5 people, and so we also put a “minus five” in the column for people.

Because our reaction produces 5 sitting people, we also put a “plus five” in the column for sitting people.

The final row of our table (the values we end with, after the reaction takes place) we fill out by simply adding the two previous rows together within each column.

So we are left with 5 people remaining, zero chairs, and 5 sitting people.

We have zero chairs remaining because chairs was the limiting reactant.

Now, let’s do another example using an actual chemical reaction.

Let's say we want to convert hydrogen and oxygen into water.

We start with 10 moles of hydrogen, 7 moles of oxygen, and we want to determine how many moles of water are produced, and which reactant was our limiting reactant.

We start by writing the balanced chemical equation for the production of water from hydrogen and oxygen

2 H2 + O2 → 2 H2O

Next, let’s write down the mole ratio for this balanced chemical reaction, using the coefficients of the two reactants on the left side of this chemical reaction:

H2/O2 = 2/1

Once again, we can pick either one of the two numbers that we are given in our problem.

Let’s say we pick the 10 moles of hydrogen. We put that in the numerator of our mole ratio equation, and then put an “x” for the number of moles of oxygen:

H2/O2 = 2/1 = 10/x

We can solve this equation using cross multiplication, which tells us that two times “x” is equal to one times ten:

(2)(x) = (1)(10)

Solving this equation tells us that x is equal to 5 moles of oxygen.

This is the amount of oxygen needed to react with 10 moles of hydrogen.

But from our problem statement, we know that we are given 7 moles of oxygen.

Since 7 is greater than 5, we know that oxygen is the reagent in excess in this reaction---we have more oxygen than we need.

This makes hydrogen our limiting reagent.

Suppose we had instead picked the other number, 7 moles of oxygen, when we set up our equation.

Writing down our equation, we would instead put our unknown variable, “y”, in place of hydrogen:

H2/O2 = 2/1 = y/7

We solve this equation again using cross multiplication, telling us that 1 times y is equal to seven times two:

(1)(y) = (7)(2)

Solving for “y” tells us that “y” equals 14 moles of hydrogen.

Since we know from our problem statement that we start out with only 10 moles of hydrogen, we can see that we have less hydrogen than we need to completely react with the oxygen.