Understanding Limiting Reactants in Chemistry

In this video, we're going to focus on limonene reactants and how to identify them. and also how to solve stoichiometry problems that involve limiting reactants. So let's go ahead and begin. Number one, zinc reacts with aqueous hydrochloric acid to produce hydrogen gas and zinc chloride. Identify the limiting reactant for each of the following situations. So let's start by writing a balanced chemical reaction. So we have zinc and its reactant with hydrochloric acid, HCl. and it's going to produce hydrogen gas which is diatomic and zinc chloride. Now what is the chemical formula of zinc chloride? Zinc has a positive 2 charge, chloride has a minus 1 charge, so therefore the chemical formula is going to be Zn1Cl2 or simply ZnCl2. Now, we need to balance this chemical reaction. And to do that, all we need to do is put a 2 in front of HCl, because we have two chlorines on the right side. After that, everything else is balanced. We have one zinc atom on both sides, and two hydrogen atoms on both sides of the equation. Now in part A, we have 12 atoms of zinc. Zinc is composed of atoms and eight molecules of HDL. Hydrochloric acid is a molecule. Which is going to be the limiting reactant? Which species? Zinc or HDL? The limiting reactant is one of the two reactants that runs out first in reaction. A quick and simple way to do that is to find which one has the lowest quantity per coefficient ratio. So we have 12 particles of zinc. The coefficient in front of it is 1. So take 12 divided by 1. The coefficient in front of HDL is 2. So 12 divided by 1 is 12. 8 divided by 2 is 4. So HDL has the lowest quantity per coefficient ratio. So therefore, HDL is the limiting reactor. It's going to run out first in this reaction. Now what about part B? We have 3 moles of zinc and 4 moles of HDL. Which one is going to run out first? Is it going to be the 3 moles of zinc or the 4 moles of HDL? Is it the one with less moles or greater moles? You can't tell just by looking at the number of moles. You're given the quantity. You need to determine the quantity per coefficient ratio. So take the number of moles and divide it by the respective coefficient. So 3 divided by 1 is 3, and 4 divided by 2 is 2. So notice that HCl has a lower quantity per coefficient ratio. So it turns out that HCl, in this example, is the limiting reactant, even though there's a greater number of moles. Oops. Now let's think about this. The ratio is 1 to 2. So 1 mole of zinc reacts with 2 moles of HDL based on the coefficients. So 2 moles of zinc will react with 4 moles of HDL. Notice that we already have 4 moles of HDL. So 4 moles of HDL can only react with 2 moles of zinc, but we have 3 moles of zinc, which means HDL is going to be gone by that time. 3 moles of zinc reacts with 6 moles of HDL. Uh-oh, we don't have 6 moles of HDL. So we're deficient in the quantity of HDL. That's why it's the limited reactant. By the time all of the zinc molecules react, we need to consume 6 moles of HDL, but we don't have that, so it's the limited reactant. But the easiest way to determine it is to find which one has the lowest quantity per coefficient ratio. Now, let's move on to Part C. We're given the grams of zinc and the grams of HDL. So, based on this, which one is going to run out first? Now, you don't want to take the mass and divide it by the coefficient. That's not going to work. The reason is because the molar mass of zinc and HDL are different. So, what you want to do is you want to convert the grams into moles. A simple way to convert grams to moles is to take the mass in grams and divide it by the molar mass. So the molar mass of zinc is 65.39 and the molar mass of HDL it's 1.008 plus 35.45 so that's about 36.458. So if we take 40 and divide it by 65.39 then that's going to give us 0.6117. moles of zinc and 56 divided by 36.458 that's 1.536 moles of HCl. Now once you have the moles of each of the two reactants then divided by the coefficients. So we're going to divide this by 1 and this one by 2. So for zinc it's still going to be 0.6117 moles per coefficient. For HCl, it's 1.536 divided by 2, which is 0.768. So in this case, zinc has the lowest quantity per coefficient ratio. So therefore, zinc is the limiting reactant in this example. So it's not because zinc has a lower mass. That's not always the case. But zinc has a much higher... molar mass than HDL, so that greatly decreased the number of moles that was present in zinc. And that's one of the reasons why zinc was the limiting reactor. The quantity of zinc was a lot less than HDL, even though we still divided HDL by 2. Number 2. Ethane reacts with oxygen gas to produce carbon dioxide and water. If you wanted to try this problem, feel free to pause the video. Let's start with a balanced chemical equation. So here's ethane. It reacts with O2. It produces CO2 and H2O. Now we have two carbons in ethane, so initially we need to put a 2 in front of CO2. There are six hydrogens on the left, so we've got to put a 3 in front of H2O, so we have six hydrogens. Now we have four oxygens from the two CO2 molecules and three from water. So that's a total of 7, which means that in order to get 7 oxygens on the left, we need a fraction, 7 over 2. Now, since we want to get rid of all fractions, I'm going to multiply everything by 2. So this now becomes 2, this is going to be 4, and this is going to be 6. 7 over 2 times 2 is 7. And now everything is balanced. So let's start with part 8. If 5 moles of ethane is placed in a container with 16 moles of oxygen gas, how many moles of CO2 will be produced? Now, if you're given a question like this, where you're given the quantity of both reactants, and you want to find the amount of product that will be produced, you need to identify the limiting reactant. There are two ways in which you could do it. First, you can do it by finding which one has the lowest quantity per coefficient ratio. like we did in part B in the last problem, or you could take the moles of each reactant, convert it to the moles of product, and find out which reactant leads to the product, well, which reactant will produce less product. The reactant that produces less product is the limited reactant. Now, for problems like this, I like to do it the second way, because I can find the amount of product produced, which is what I want to do in this problem, and at the same time, identify the limiting reactant. But for this example, part A, I'm going to do it both ways. So let's identify the limiting reactant the old-fashioned way. So we're going to divide the moles of ethane by 2 and the moles of O2 by its coefficient 7. So 5 divided by 2 is 2.5, and 16 divided by 7, I need to calculate for that, that's about 2.29. So 2.29 is less than 2.5. So O2 has a lower mole per coefficient ratio, which means O2 in this example is the limited reactant. Now, because O2 is a limited reactant, it's going to produce a lower amount of product. So it's going to yield less moles of CO2. But I'm going to show it to you. We're going to do it both ways. Let's convert the moles of ethane. to the moles of CO2 first. So we have 5 moles of C2H6. And the molar ratio is 2 to 4. So for every 2 moles of C2H6 that reacts, 4 moles of CO2 will be produced. So 5 times 4 is 20, and 20 divided by 2 is 10. So 10 moles of CO2 can be produced if all of the 5 moles of ethane participates in this reaction. Now let's start with the 16 moles of O2. Even though the moles of O2 is much greater than the moles of ethane, O2 was still the limited reactant due to the high coefficient that's in front of it. So now, to convert it from O2 to CO2, we need to use the molar ratio between O2 and CO2, which is 7 to 4. So for every 7 moles of O2 that reacts, 4 moles of CO2 will be produced. 16 times 4 is 64, and 64 divided by 7 is 9.14. So as you can see, this is the theoretical yield of the reaction. That's the theoretical yield of CO2. The theoretical yield is the maximum amount of product that you can get in a reaction. It's always going to be the smaller of the two values. So the answer for part A is 9.14. Now the reactant that gave you the lower theoretical yield, that's O2, the limiting reactant. So, that's why sometimes it might be better just to take each reaction, excuse me, each reactant and see how much product will be produced from each reactant. The reactant that produces less product will be the limited reactant and the amount of product that you get from the limited reactant is the theoretical yield or the amount of CO2 that will be produced in a reaction. So now let's move on to our next part. Now I need to get rid of this as well. Now keep in mind, we don't know if O2 is going to be the limiting reactant in this second part, because we have new values. We're given 30 grams of ethane and 84 grams of O2. Now, if we wanted to find the limiting reactant the old-fashioned way, we would have to convert grams to moles and divide it by the coefficients. Instead of doing all that, let's just take each reactant, convert it to the grams of product, see which one produces less product. The reactant that does so is limited reactant. And the amount of product that is less than the other one, that's going to be the actual amount of water that will be produced. So it's like catching two birds with a single stone. Let's start with ethane. Let's convert 30 grams of ethane to grams of water. Now keep in mind to perform a gram to gram conversion is a three-step process. First, You need to take the grams of substance A and change it to the moles of substance A using the molar mass of A. Next, you need to use the molar ratio to convert from substance A to B while keeping the unit the same. And in the third step, you need to change the unit from moles to grams. So that's how you can perform a gram-to-gram conversion. So first, we need to find the molar mass of ethane. So we have two carbons and six hydrogen atoms. So it's 2 times 12.01 plus 6 times 1.008. So that's about 30.068 grams per mole. So let's put the grams on the bottom and the moles of ethane on top. Now let's use the molar ratio to change the substance from ethane to water. So it's a 2 to 6 ratio. For every 2 moles of ethane that reacts, 6 moles of water will be produced. Now finally, let's convert to grams of water. So we have two hydrogen atoms, which is 2 times 1.008, plus the atomic mass of oxygen, which is 16. So that will give us a molar mass of 18.016 grams per mole. So that's how you can perform a gram-to-gram conversion. So it's going to be 30 divided by 30.068. times 6 divided by 2 times 18.016 so the answer that we have from ethane is 53.93 grams so that's from ethane i'm going to highlight it in red Now let's do the same using oxygen. Let's start with 84 grams of O2 and let's convert it to the grams of water. So the molar mass of O2 is going to be 2 times 16, which is 32. Now let's change the substance using the molar ratio. The molar ratio between O2 and water is 7 to 6. So for every... Every 6 moles of water that's produced, 7 moles of O2 participates in the reaction. And then let's convert moles to grams. We already have the molar mass of H2O. That's 18.016 grams. So let's do the math. It's 84 divided by 32 times 6 divided by 7 times 18.016. So the answer that I have is 40.54 grams of water. Now, which one is the correct value? So, what's the answer to this question? How many grams of water will be produced in this reaction? Is it 53.9 or 40.5? It's going to be the smaller of the two values. It's going to be 40.5. That's the answer to part B. The reactant that gave us the lower theoretical yield of 40.5 grams of water is O2. So once again, O2 was... the limited reactant which means that ethane is the excess reactant because it gave us too much product so by the time also runs out the reaction stops so o2 is the limited reactant it determines the grams of product product that is produced, which is the theoretical yield. The theoretical yield could be the moles of product or the grams of product. It's simply the maximum quantity of product that can be formed in a reaction. So what this means is that if all of the 84 grams of O2 reacts, which it will because it's a limited reactant, only 40.5 grams of water will be produced. Now if ethane was a limited reactant, if all of the 30 grams of ethane reacted, which it reaction, 53.9 grams of water could be produced. That's if we had a lot of oxygen in the air. But in this example, oxygen is a limited reactant. So the amount of oxygen determines how many grams of water will be produced in this reaction. So, keep that in mind. If you want to take the grams of each reactant, convert it to the grams of product, the theoretical yield is the one with the lower value. And the reactant that gave you the lower grams of product is the limiting reactant.

In this video, we're going to focus on limonene reactants and how to identify them. and also how to solve stoichiometry problems that involve limiting reactants. So let's go ahead and begin.

Number one, zinc reacts with aqueous hydrochloric acid to produce hydrogen gas and zinc chloride. Identify the limiting reactant for each of the following situations. So let's start by writing a balanced chemical reaction.

So we have zinc and its reactant with hydrochloric acid, HCl. and it's going to produce hydrogen gas which is diatomic and zinc chloride. Now what is the chemical formula of zinc chloride? Zinc has a positive 2 charge, chloride has a minus 1 charge, so therefore the chemical formula is going to be Zn1Cl2 or simply ZnCl2. Now, we need to balance this chemical reaction.

And to do that, all we need to do is put a 2 in front of HCl, because we have two chlorines on the right side. After that, everything else is balanced. We have one zinc atom on both sides, and two hydrogen atoms on both sides of the equation.

Now in part A, we have 12 atoms of zinc. Zinc is composed of atoms and eight molecules of HDL. Hydrochloric acid is a molecule.

Which is going to be the limiting reactant? Which species? Zinc or HDL? The limiting reactant is one of the two reactants that runs out first in reaction. A quick and simple way to do that is to find which one has the lowest quantity per coefficient ratio.

So we have 12 particles of zinc. The coefficient in front of it is 1. So take 12 divided by 1. The coefficient in front of HDL is 2. So 12 divided by 1 is 12. 8 divided by 2 is 4. So HDL has the lowest quantity per coefficient ratio. So therefore, HDL is the limiting reactor. It's going to run out first in this reaction.

Now what about part B? We have 3 moles of zinc and 4 moles of HDL. Which one is going to run out first?

Is it going to be the 3 moles of zinc or the 4 moles of HDL? Is it the one with less moles or greater moles? You can't tell just by looking at the number of moles. You're given the quantity. You need to determine the quantity per coefficient ratio.

So take the number of moles and divide it by the respective coefficient. So 3 divided by 1 is 3, and 4 divided by 2 is 2. So notice that HCl has a lower quantity per coefficient ratio. So it turns out that HCl, in this example, is the limiting reactant, even though there's a greater number of moles. Oops.

Now let's think about this. The ratio is 1 to 2. So 1 mole of zinc reacts with 2 moles of HDL based on the coefficients. So 2 moles of zinc will react with 4 moles of HDL. Notice that we already have 4 moles of HDL.

So 4 moles of HDL can only react with 2 moles of zinc, but we have 3 moles of zinc, which means HDL is going to be gone by that time. 3 moles of zinc reacts with 6 moles of HDL. Uh-oh, we don't have 6 moles of HDL.

So we're deficient in the quantity of HDL. That's why it's the limited reactant. By the time all of the zinc molecules react, we need to consume 6 moles of HDL, but we don't have that, so it's the limited reactant. But the easiest way to determine it is to find which one has the lowest quantity per coefficient ratio.

Now, let's move on to Part C. We're given the grams of zinc and the grams of HDL. So, based on this, which one is going to run out first? Now, you don't want to take the mass and divide it by the coefficient. That's not going to work.

The reason is because the molar mass of zinc and HDL are different. So, what you want to do is you want to convert the grams into moles. A simple way to convert grams to moles is to take the mass in grams and divide it by the molar mass.

So the molar mass of zinc is 65.39 and the molar mass of HDL it's 1.008 plus 35.45 so that's about 36.458. So if we take 40 and divide it by 65.39 then that's going to give us 0.6117. moles of zinc and 56 divided by 36.458 that's 1.536 moles of HCl.

Now once you have the moles of each of the two reactants then divided by the coefficients. So we're going to divide this by 1 and this one by 2. So for zinc it's still going to be 0.6117 moles per coefficient. For HCl, it's 1.536 divided by 2, which is 0.768. So in this case, zinc has the lowest quantity per coefficient ratio. So therefore, zinc is the limiting reactant in this example.

So it's not because zinc has a lower mass. That's not always the case. But zinc has a much higher... molar mass than HDL, so that greatly decreased the number of moles that was present in zinc. And that's one of the reasons why zinc was the limiting reactor.

The quantity of zinc was a lot less than HDL, even though we still divided HDL by 2. Number 2. Ethane reacts with oxygen gas to produce carbon dioxide and water. If you wanted to try this problem, feel free to pause the video. Let's start with a balanced chemical equation.

So here's ethane. It reacts with O2. It produces CO2 and H2O. Now we have two carbons in ethane, so initially we need to put a 2 in front of CO2. There are six hydrogens on the left, so we've got to put a 3 in front of H2O, so we have six hydrogens.

Now we have four oxygens from the two CO2 molecules and three from water. So that's a total of 7, which means that in order to get 7 oxygens on the left, we need a fraction, 7 over 2. Now, since we want to get rid of all fractions, I'm going to multiply everything by 2. So this now becomes 2, this is going to be 4, and this is going to be 6. 7 over 2 times 2 is 7. And now everything is balanced. So let's start with part 8. If 5 moles of ethane is placed in a container with 16 moles of oxygen gas, how many moles of CO2 will be produced? Now, if you're given a question like this, where you're given the quantity of both reactants, and you want to find the amount of product that will be produced, you need to identify the limiting reactant. There are two ways in which you could do it.

First, you can do it by finding which one has the lowest quantity per coefficient ratio. like we did in part B in the last problem, or you could take the moles of each reactant, convert it to the moles of product, and find out which reactant leads to the product, well, which reactant will produce less product. The reactant that produces less product is the limited reactant.

Now, for problems like this, I like to do it the second way, because I can find the amount of product produced, which is what I want to do in this problem, and at the same time, identify the limiting reactant. But for this example, part A, I'm going to do it both ways. So let's identify the limiting reactant the old-fashioned way.

So we're going to divide the moles of ethane by 2 and the moles of O2 by its coefficient 7. So 5 divided by 2 is 2.5, and 16 divided by 7, I need to calculate for that, that's about 2.29. So 2.29 is less than 2.5. So O2 has a lower mole per coefficient ratio, which means O2 in this example is the limited reactant. Now, because O2 is a limited reactant, it's going to produce a lower amount of product.

So it's going to yield less moles of CO2. But I'm going to show it to you. We're going to do it both ways.

Let's convert the moles of ethane. to the moles of CO2 first. So we have 5 moles of C2H6. And the molar ratio is 2 to 4. So for every 2 moles of C2H6 that reacts, 4 moles of CO2 will be produced.

So 5 times 4 is 20, and 20 divided by 2 is 10. So 10 moles of CO2 can be produced if all of the 5 moles of ethane participates in this reaction. Now let's start with the 16 moles of O2. Even though the moles of O2 is much greater than the moles of ethane, O2 was still the limited reactant due to the high coefficient that's in front of it.

So now, to convert it from O2 to CO2, we need to use the molar ratio between O2 and CO2, which is 7 to 4. So for every 7 moles of O2 that reacts, 4 moles of CO2 will be produced. 16 times 4 is 64, and 64 divided by 7 is 9.14. So as you can see, this is the theoretical yield of the reaction. That's the theoretical yield of CO2. The theoretical yield is the maximum amount of product that you can get in a reaction.

It's always going to be the smaller of the two values. So the answer for part A is 9.14. Now the reactant that gave you the lower theoretical yield, that's O2, the limiting reactant.

So, that's why sometimes it might be better just to take each reaction, excuse me, each reactant and see how much product will be produced from each reactant. The reactant that produces less product will be the limited reactant and the amount of product that you get from the limited reactant is the theoretical yield or the amount of CO2 that will be produced in a reaction. So now let's move on to our next part.

Now I need to get rid of this as well. Now keep in mind, we don't know if O2 is going to be the limiting reactant in this second part, because we have new values. We're given 30 grams of ethane and 84 grams of O2. Now, if we wanted to find the limiting reactant the old-fashioned way, we would have to convert grams to moles and divide it by the coefficients.

Instead of doing all that, let's just take each reactant, convert it to the grams of product, see which one produces less product. The reactant that does so is limited reactant. And the amount of product that is less than the other one, that's going to be the actual amount of water that will be produced.

So it's like catching two birds with a single stone. Let's start with ethane. Let's convert 30 grams of ethane to grams of water.

Now keep in mind to perform a gram to gram conversion is a three-step process. First, You need to take the grams of substance A and change it to the moles of substance A using the molar mass of A. Next, you need to use the molar ratio to convert from substance A to B while keeping the unit the same.

And in the third step, you need to change the unit from moles to grams. So that's how you can perform a gram-to-gram conversion. So first, we need to find the molar mass of ethane. So we have two carbons and six hydrogen atoms. So it's 2 times 12.01 plus 6 times 1.008.

So that's about 30.068 grams per mole. So let's put the grams on the bottom and the moles of ethane on top. Now let's use the molar ratio to change the substance from ethane to water.

So it's a 2 to 6 ratio. For every 2 moles of ethane that reacts, 6 moles of water will be produced. Now finally, let's convert to grams of water.

So we have two hydrogen atoms, which is 2 times 1.008, plus the atomic mass of oxygen, which is 16. So that will give us a molar mass of 18.016 grams per mole. So that's how you can perform a gram-to-gram conversion. So it's going to be 30 divided by 30.068. times 6 divided by 2 times 18.016 so the answer that we have from ethane is 53.93 grams so that's from ethane i'm going to highlight it in red Now let's do the same using oxygen. Let's start with 84 grams of O2 and let's convert it to the grams of water.

So the molar mass of O2 is going to be 2 times 16, which is 32. Now let's change the substance using the molar ratio. The molar ratio between O2 and water is 7 to 6. So for every... Every 6 moles of water that's produced, 7 moles of O2 participates in the reaction.

And then let's convert moles to grams. We already have the molar mass of H2O. That's 18.016 grams.

So let's do the math. It's 84 divided by 32 times 6 divided by 7 times 18.016. So the answer that I have is 40.54 grams of water.

Now, which one is the correct value? So, what's the answer to this question? How many grams of water will be produced in this reaction?

Is it 53.9 or 40.5? It's going to be the smaller of the two values. It's going to be 40.5.

That's the answer to part B. The reactant that gave us the lower theoretical yield of 40.5 grams of water is O2. So once again, O2 was...

the limited reactant which means that ethane is the excess reactant because it gave us too much product so by the time also runs out the reaction stops so o2 is the limited reactant it determines the grams of product product that is produced, which is the theoretical yield. The theoretical yield could be the moles of product or the grams of product. It's simply the maximum quantity of product that can be formed in a reaction.

So what this means is that if all of the 84 grams of O2 reacts, which it will because it's a limited reactant, only 40.5 grams of water will be produced. Now if ethane was a limited reactant, if all of the 30 grams of ethane reacted, which it reaction, 53.9 grams of water could be produced. That's if we had a lot of oxygen in the air.

But in this example, oxygen is a limited reactant. So the amount of oxygen determines how many grams of water will be produced in this reaction. So, keep that in mind. If you want to take the grams of each reactant, convert it to the grams of product, the theoretical yield is the one with the lower value.

And the reactant that gave you the lower grams of product is the limiting reactant.

Transcript for:Understanding Limiting Reactants in Chemistry

Transcript for:
Understanding Limiting Reactants in Chemistry