Limiting reactants and percent yield
Definitions & Key takeaways
A limiting reactant in a chemical reaction is the reactant that's used up first, so, it's the one that limits the amount of products that are created in a chemical reaction. The other reactants are then said to be in excess. To get the most out of a chemical reaction, it's important to use the correct amounts of all the reactants. If you have too much of one reactant and not enough of another, you won't get the maximum yield.
The percent yield is a measure of how efficient a chemical reaction is. It's calculated by comparing the amount of product that's actually produced to the maximum amount that could be produced under perfect conditions. A percent yield below 100% means there was some loss of product during the reaction. This can happen due to incomplete mixing or because some reactants are used up before others.
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 in five 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.
One person plus one chair equals one sitting person. So because it takes one person and one chair to produce one sitting person that we know that with five chairs in 10 people, we are limited to producing at most five sitting people.
So the number of chairs is the limiting reactant or limiting reagent in our example. But sometimes figuring out the lining reactant is not very intuitive.
And so we use the chair example to learn how to find the lining 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 1 to 1. Now that we have the mole ratio between the reactants, we pick either one of the numbers that were given in the problem.
We can pick either five chairs or we can pick 10 people. It doesn't matter here.
Suppose we pick 10 people. We write this number into our mole ratio equation and then put an X in the place of the number of chairs.
First thing divided by chair equals one, divided by one equals 10, divided by X. If we solve the equation, we would find that X is equal to 10 and 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 five 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 five chairs. In this case, we would write out the mole ratio with an unknown Y in place of the number of people person divided by chair equals one, divided by one equals Y divided by five.
We can solve this equation in order to get Y equals five, meaning that we would need exactly five people to react with five 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'll 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, five chairs and zero sitting people. We just determined that chairs were our limiting reactant.
And so all the chairs get used up in the reaction. So in the change row, which is what happens during the reaction, we put a minus five in the column for chairs, these chairs reacted with five people.
And so we also put a minus five in the column for people because our reaction produces five sitting people. We also put a plus five in the column for sitting people.
The final row of our table, which is the value we end with. After the reaction takes place, we fill out by simply adding the two previous rows together within each column.
So we're left with five people remaining zero chairs and five sitting people. We have zero chairs remaining because the chairs were the limiting reactant.
Now, let's do another example, using an actual chemical reaction. Let's say we wanna convert hydrogen and oxygen into water.
We start with 10 moles of hydrogen and seven 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.
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 the chemical equation H two divided by O2 equals two, divided by one.
Once again, we can pick either one of the two numbers that we are given in the 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 in X for the number of moles of oxygen, we can solve this equation using cross multiplication which tells us that two times X is equal to one times 10.
Solving this equation tells us that X is equal to five 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 seven moles of oxygen since seven is greater than five, we know that oxygen is the reagent in excess in this reaction.
In other words, we have more oxygen than we need. This makes hydrogen our limiting reagent.
Suppose we had instead picked the other number seven moles of oxygen. When we set up our equation, writing down our equation, we would instead put our unknown variable.
Why in place of hydrogen, we solve this equation again using cross multiplication, telling us that one times Y is equal to seven times two.
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 oxygen.
So we once again find that hydrogen is our limiting reactant. Once you identify the limiting reactant, you can use the limiting reactant and the mole ratio to figure out how many moles of your product you'll make.
Once again, the easiest way to do this is to draw an ICF table. The columns in our table, this time will be hydrogen, oxygen and water.
In the first row row, I, we put down the starting conditions before our reaction takes place. 10 moles of hydrogen, seven moles of oxygen and zero moles of water.
In the second row row C, we put down the changes that occurred during our reaction because we just found that hydrogen is our limiting reactant and we'll use it all up in the reaction.
We put a minus 10 in that row. Since the mole ratio of hydrogen to oxygen is 2 to 1, 10 moles of hydrogen will use up five moles of oxygen.
So we put a minus five in the row for oxygen. Since the limiting reactant limits how much product we make.
We look at the mole ratio of the limiting reactant to our product. Since the mole ratio of hydrogen to water is 2 to 2, 10 moles of hydrogen will make 10 moles of water.
So we put a plus 10 in the row for water. In the third row row F, we calculate the final number of moles of each chemical and the reaction adding everything together.
We're left with two moles of leftover oxygen, 10 moles of water and zero moles of hydrogen remaining, which makes sense because hydrogen is our limiting reagent.
It's important to remember that once the limiting reactant is used up, the reaction will stop. We can visualize this concept by looking at a diagram of the atoms and molecules participating in our reaction.
First, we can visualize the starting conditions for our reaction, which is a mixture of gaseous molecules of hydrogen and oxygen, which have not undergone any chemical reaction.
We can count these molecules and see that we have 10 molecules of hydrogen and seven molecules of oxygen. Next, we can look at our molecules again, after the reaction has taken place.
If we count again, we see that we have 10 molecules of water, zero molecules of hydrogen and two remaining molecules of oxygen.
So our limiting reactant hydrogen got totally used up, but we do have some leftover oxygen. Let's do another limiting reactant example.
Suppose we have some solid zinc which reacts with aqueous silver nitrate. This reaction produces two products, solid silver and aqueous zinc nitrate.
Supposed we start with 2.5 g of zinc and 2.5 g is silver nitrate. And we're asked to find the limiting reactant as well as the amount of silver metal produced.
First, we need to find the starting number of moles of zinc. And we do that by dividing the grams by the molar mass of zinc, we can look up the molar mass on the periodic table and it's 65.38 g per mole.
So at 2.50 g divided by 65.38 g per mole gives us 0.0382 moles of zinc. Next, we need to find the starting number of moles of silver nitrate.
Again, we look it up adding each element's molar mass together and the total is 100 and 69.9 g per mole. Using the same method.
We see 2.50 g divided by 100 and 69.9 g per mole gives us 0.0147 moles of silver nitrate. Next, we write out our balance chemical equation.
Once again, we can start by writing down the mole ratio for this balanced chemical reaction using the coefficients of the two reactants on the left side of this chemical reaction.
Our next step is to pick either one of the two starting amounts of the reactants in our problem. Let's pick 0.0382 moles of zinc.
We write down the mole ratio equation again. When we put an X in the place of the starting moles of silver nitrate, we can solve this equation for X using cross multiplication which tells us that X equals 0.0764 moles of silver nitrate.
This is the amount of silver nitrate needed to react with the zinc that we have. However, we know from our earlier calculations that we only start out with 0.0147 moles of silver nitrate.
Since we have less silver nitrate than we need. Silver nitrate is our limiting reactant.
We're now ready to figure out the massive silver that our reaction produces. Based on the chemical reaction equation.
Two moles of silver are produced for every two moles of silver nitrate consumed. Since silver nitrate as our limiting reactant, we know that all of it is consumed all 0.0147 moles.
And since the mole ratio is 2 to 2, that means that 0.0147 moles of silver must be produced by a reaction. To find the total mass of silver produced.
We need to consult the periodic table. Again, the molar mass of silver is 100 and 7.9 g per mole.
We multiply 0.0147 moles by 100 and 7.87 g per mole. To find that 1.59 g of silver are produced 1.59 g of silver is the theoretical yield of our reaction.
In other words, it's the yield of product that would be produced if the reaction proceeds perfectly. However, in many real world cases, the reaction produces less product than the theoretical yield.
For example, suppose that we performed our reaction in a laboratory and we only measured an output of 1.00 g of silver. We could call this number the actual yield of our reaction.
Now, we can calculate the percent yield. In order to measure the overall efficiency of our laboratory process, we simply divide the actual yield by the theoretical yield and then multiply the result by 100.
So at 1.00 divided by 1.59 times 100 gives a percent yield of 62.9%. If we want to produce silver from silver nitrate with higher efficiency, then we'd want to get the percent yield as close to 100% as possible.
All right, as a quick recap limiting reactants limit the amount of products that are created in the chemical reaction during a reaction.
Theoretically, the entire quantity of limiting reagent is consumed, determining the final amount of product and thus the theoretical yield.
However, in real world applications, the actual yield of a reaction is lower than the theoretical yield. We can measure the efficiency of a reaction using the percent yield, which compares the actual yield to the theoretical yield.
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