Hello, my name's Mrs. Niven, and today we're going to be talking about limiting reactants as part of our unit on calculations involving masses.

Now, you may have come across this phrase of a limiting factor from some of your previous learning, but what we do in today's lesson will help us to not only answer that big question of what are substances made of, but what also help us to better appreciate the interaction between particles in a chemical reaction and the amount of product that might form from them.

It will also help us to better consider as well the amounts of reactants that we're starting with so that we as chemists might be able to consider how we can adjust those so that our reactions are less wasteful going forward.

So by the end of today's lesson, you should feel more comfortable being able to identify a reactant as either being in excess or limiting and use that information then to calculate the theoretical yield for a reaction.

Now throughout the lesson I will be referring to some key words, and these include limiting reactant, excess reactant, theoretical yield, mole, and stoichiometry.

Now the definitions for these keywords are given in sentence form on the next slide, and you may wish to pause the video here so you can read through them or maybe even take a quick note of them so you can refer back to it later on in your lesson or later on in your learning.

So today's lesson is broken into two parts.

Firstly, we'll look at how we can distinguish between a limiting and an excess reactant, and then we'll move on to look at how we can use that information to calculate a theoretical yield.

So let's get started by looking at the difference between a limiting and excess reactant.

Now, you may have heard the phrase that chemistry is very much like cooking, and believe it or not, it actually is.

I mean, recipes and reaction equations have quite a lot in common.

For instance, they tell us what we might need and how much of what we'll need.

They also tell us what we'll be able to make and how much we'll be able to make.

Now in a chemical reaction what's actually happening are those reactant atoms rearranging to form the products, and because of that, those atoms and the mass of those atoms are conserved throughout the reaction.

And our balanced symbol equations then show that conservation using coefficients, those large numbers that come before the formula, and we can interpret then those coefficients as representing the atoms, a molecule or even a formula unit.

What's interesting if you take a closer look at a cooking recipe is that it's not just indicating the ingredients you need or how much, what it's really doing is indicating the ratio of the ingredients that are needed.

So for instance, here I have a recipe to make some six pancakes and it tells me the amounts of each ingredients as well.

And what I can do with this recipe then is if I want to now go from making six pancakes to 12 pancakes, I can use the ratio of that original recipe to make those 12 pancakes because 12 is simply twice, two times, six.

So what I do then is simply take the recipe for my six pancakes, multiply everything in there by two, and I now have a recipe to make 12 pancakes simply by maintaining the ratio of that original recipe.

Now, the ratios between different substances in a chemical reaction are indicated by the stoichiometry of a balanced symbol equation.

That stoichiometry or the molar ratio is actually shown by the coefficients of that balanced symbol equation.

Now if we remember that a mole is representing 6.

02 times 10 to the 23 particles of any substance, we can then interpret the coefficients of a balanced symbol equation to tell us a little bit about the number of particles that are present and the ratio of those particles within a chemical reaction.

For instance, if I look at this reaction of magnesium reacting with oxygen, I can use those coefficients to actually say, I have two moles of magnesium.

I'm going to need at least one mole of oxygen molecules in order for them to react and form them two moles of formula units of our magnesium oxide.

So similar to my cooking recipe, the stoichiometry or the molar ratio for a chemical reaction doesn't change even if the quantitative amounts change.

So the stoichiometry shows me my recipe for that chemical reaction, that molar ratio, how much of each ingredient do I need in order to make a certain amount of product.

And just like in my cooking recipe, if I want to change the amount of product or if I'm changing the amount of my starting materials, then that ratio needs to be maintained.

So for instance, if I started now with twice as much magnesium, I'm going to need to multiply the amount of oxygen that I need for that reaction to take place.

And then I can also say that I have multiplied the amount of product that could possibly form to maintain that ratio.

Similarly, if I use half as much magnesium, I would need half as much oxygen and I would make half as much magnesium oxide, maintaining that ratio.

Let's stop here for a quick check.

Which chemical equation best describes the ratios shown in the diagram of the chemical reaction below? Now, you may wish to pause the video here to discuss your ideas and then come back when you're ready to check your answer.

Now, the first thing I would've done is actually tried to label up what the different molecules are using that key, that the X is represented by the black circles and Y is represented by the white circles.

So once I've labelled up my reactants and my products, hopefully and very well done if you have chosen C to show the ratios, there might have been a temptation to choose A as an option, but that's showing the actual amounts that are shown in the particle diagrams. What we wanted was the ratio, and because of that we're looking for those lowest possible coefficients.

So very well done if you chose C.

Great start guys, well done.

Now because of these required ratios, there can sometimes then be a limit to the amount of product that can be made.

So if we go back to this idea of our recipe, so making pancakes, and this particular recipe requires two eggs, but I only have one egg.

So that means that I'm limited in the amount of pancakes I can make because of the number of eggs that I have available.

I have half as many eggs, therefore my pancake recipe is going to change.

In order to maintain that ratio, I need to divide all of my ingredients by two in order to create a new pancake recipe that is limited by the number of eggs that I have available.

And because of that, I'm not going to be able to make the same amount of pancakes.

My recipe has changed, I have less batter and I have half as much in fact, and therefore I'm going to make half as much product.

So a substance that is limiting the amount of product that can form is referred to as the limiting reactant.

It has limited how much you can make, it's restricted that.

Let's stop for another quick check.

A sausage sandwich is made from one sausage and one bun.

What do you think is going to be the limiting reactant to make a sausage sandwich given the ingredients that are shown below? And if you figured that out, why don't you challenge yourself? How many sausage sandwiches could you make Well done if you chose B.

The sausages are my limiting reactant to make a sausage sandwich, and because of that I will only be able to make three sausage sandwiches using the ingredients that are shown.

So very well done if you got that correct.

Now, if we have one reactant that is limiting, there is the possibility that we have another substance that is now going to be in surplus.

It's extra to what we actually need in order to make as much product as possible.

And because of that, we refer to those surplus or those extra reactants as being in excess, excessive, being too much.

I have too much of it.

So we already said that the sausages were my limiting reactant in my sausage sandwich scenario.

That means then the hotdog buns are my excess reactant, and that's because there are going to be some buns that are left over.

They are unused.

Now the thing to remember about an excess reactant particle is that they will be present in the final reaction mixture, but we don't include them in the chemical equation.

If we take a closer look at this particle diagram that I've drawn, I know that the blue substance is my excess reactant and the white substance is the limiting reactant because all of those white particles have been rearranged and used to make a product.

But I have some blue particles on the product side of my reactant that look identical to what they were like on the reactant side.

They have not changed, they've not rearranged in any way, shape or form.

So what I started with as a reactant is still present in my product final mixture.

And because of that, they've not reacted, I have extra excess particles.

So why aren't these included in a chemical equation? Well, we have to remember that a chemical equation is showing us that stoichiometric or molar ratio relationship between the particles involved in a reaction.

How much or how many particles do I need of each reactant.

And then indicating how much or how many particles of product I can form from a reaction involving that number of particles.

It's my recipe, not showing me a specific situation.

If I use my recipe with different numbers or amounts of reactants, I could have potentially a different limiting or excess reactant.

So our balanced symbol equation acts as our recipe, our starting point, and then we need to examine each situation individually to decide, okay, well in this scenario, which will be my limiting and which will be my excess reactant.

Let's stop for another quick check.

A reaction is shown by the balanced symbol equation of X two plus Y two makes two XY.

What I'd like you to do is to complete the diagram below to show the molecules that will be present when that reaction has finished.

And the X particles are represented by the black circles and the Y particles are represented by the white circles.

And if you finished your diagram, think about which reactant do you think was the limiting reactant and which do you think was in excess.

Now this might take a little bit of time, so definitely pause the video and come back when you're ready to check your answers.

Okay, let's see how you got on.

So if you've drawn your diagram correctly, it should look a little bit like this.

You should have four particles that are represented by a black and a white circle, and then two particles that are shown as two black particles together.

And because of that, the excess reactant then was the X substance and the limiting reactant was the Y substance because all of those particles had rearranged in the products.

A very well done if you've got your diagram correct and incredibly well done if you managed to correctly identify the excess and limiting reactants here.

Great job guys.

Okay, time for our first task.

What I'd like you to do is to complete the Venn diagram by placing the statements that are provided around the Venn diagram into the appropriate area of it.

So it might be in the circles for limiting or excess, or it might be in the overlap region, but do beware of any red herrings in those statements.

So pause the video and come back when you're ready to check your work.

Okay, let's see how you got on.

Now, before we go through this, I just want to take a moment to appreciate that some people can find completing a Venn diagram particularly tricky and might actually try to go outta their way to avoid it, and I appreciate that.

But when it comes to chemistry, we use key terms that are sometimes very similar but have subtle differences between them and these are really useful tasks to help us to really identify those similarities and differences.

So we can use those terms correctly going forward in our descriptions, but also in identifying different parts of chemistry that's going forward.

So in this particular one, I always start with my Venn diagrams, looking at what's the similarity, what's linking these ideas in the first place? And for limiting and excess, we're talking about reactants only and we're talking about something that's composed of particles.

And then it's the differences that are gonna help me going forward to use these terms correctly.

A limiting reactant is used up in a reaction and the amount of product formed is going to depend on them.

An excess reactant on the other hand is left over in a reaction, and because of that, it is still present in my reaction mixture at the end, and that's important for chemists to appreciate because you might need to then consider a separation technique to isolate that reaction product from that excess reactant.

So it's really important that we understand the differences between limiting and excess.

The red herrings then for this was the amount of reactant depends on this and product.

So very well done if you manage to complete your Venn diagram correctly guys.

What a great start.

For the next part of this task, I'd like you to consider the sausage sandwiches again, but the eagle eye among you may have noticed that the sausages here are a little bit thinner than the last ones.

Now because of that we're going to use one bun and two sausages.

So I'd like you to consider that in this scenario, what do you think the limiting reactant might be, and then how many sausage sandwiches can actually be made using those ingredients shown? So pause the video and come back when you're ready to check your answers.

Okay, let's see how you got on.

Now we can see from the description of the picture that we had 12 sausages and 10 buns on that grill.

And in this scenario, the sausages are going to be limiting because all 12 of them will be used up forming the sausage sandwich that's made of one bun and two sausages.

So how many sausage sandwiches are we actually making then using those ingredients? Well, because the sausages are limiting, we can only make six sausage sandwiches even though we have 10 buns on the grill.

So very well done if you manage to get those correct.

Okay, for the last part of this task, I'd like you to consider the chemical reaction that's represented by the balanced chemical equation of X two plus Y two makes two XY.

And I'd like you to use the key that's shown to draw the molecules that will be present at the end of each reaction.

So pause the video and come back when you're ready to check your work.

Okay, let's see how you got on.

So what you needed to do was to use the key that was provided and the ratios that are shown in your balanced chemical equation to draw your diagrams. And for A, it should look something like this.

And when I compare my key to the diagrams that I have, I actually have a Y two plus X two makes two XY, which is very similar to what my balance chemical equation was to start with.

If I look at B then, I can actually see that I have twice as much of my reactants, so I should be able to make twice as much of my products showing a diagram that looks like this.

Moving on to letter C, your diagram should look a little bit like this.

And for D, your diagram will look like this.

Now notice that in both C and D there are molecules present at the end of the reaction that look the same as our reactants, so we have some excess particles present.

It's important that you have included those in your diagram at the end, okay? So don't forget to have included those and if you have incredibly well done, really pleased with the progress you guys are making in this lesson, guys.

Keep it up.

Now that we're feeling more confident distinguishing between a limiting and excess reactant, let's look at how we can use that information to calculate a theoretical yield.

Now, we need to remember that chemists use mathematical relationships in order to calculate the number of particles that are present in a sample of a substance, and that can include the mass of a reactant.

And that we refer to the number of particles in a sample as the number of moles.

So the moles are calculated by taking the massing grammes of our substance and dividing by that relative mass of the substance we're looking at.

Now, what that actually means then is the number of moles represents the number of particles that are available to react when we're looking at our reactants.

And from this information, chemists can then determine which of our reactants is going to be limiting and which is going to be in excess.

So if we look at this particle diagram, we can see then the limiting reactant is the blue particles because they've all been rearranged into the products and that the red particles are going to be the excess reactant because on the product side of our equation here, they are unchanged.

They have not rearranged or reorganised to create a product, and that's all because of looking at the particles that were available to react in the first place.

Let's stop here for a quick check.

Which reactant is the limiting reactant in this example? And I'd like you to explain your choice.

So I'm looking for a because clause.

You may wish to pause the video and come back when you're ready to check your work.

Well done if you said X two was the limiting reactant.

Now I've said X two because when we look at the diagram for our reactants, we can see that everything is a diatomic molecule, but still give yourself a mark if you simply said X.

Now that is the limiting reactant because all of its particles have reacted or rearranged in order to form the product, and we know that because there are no excess X two particles shown in that final reaction mixture.

So very well done if you managed to get that correct, guys, great job.

Now, once a limiting reactant has been identified for a particular situation, the chemists can then calculate the maximum theoretical yield, or theoretically, how much product could we make for a reaction by exploiting that stoichiometry or the molar ratios of a chemical equation.

Now, a theoretical yield assumes that all of those limiting reactant particles have reacted, and formed as much product as possible.

So in this particle diagram, we can identify that limiting reactant as being the blue particles 'cause all of them in this diagram have reacted, and that means then our maximum theoretical yield has assumed that every single one of those particles have reacted.

Now, the thing to remember about a theoretical yield is that they are always calculated and never measured.

Okay, so a theoretical yield is based on the calculations that you're doing with your calculator and a different type of yield known as the actual yield is what you might measure in the laboratory.

Now, in order to calculate a theoretical yield for a reaction, you're going to need a few things.

You'll need to know the number of moles for all the reactants in that reaction.

You'll need a calculator, a balanced symbol equation, so you have that stoichiometry or the molar ratio between the materials in that reaction, and you'll need a step-by-step strategy to follow, which I'll go through in a moment.

But the reason we need to start with the number of moles in a substance is that we need to know the number of particles that are available to react in order to determine what is our limiting reactant to start with before we can then calculate a yield for that reaction.

So that's why we need to know the number of moles to start with.

What I'll do now then is go through a strategy that you can use in order to identify the limiting reactant for a particular scenario and then use that information to calculate a theoretical yield.

We'll start by looking at just moles, so number of particles to start with, and then later on we'll extend to look at including masses.

So in this first example, we want to know how many moles of aluminium chloride can be produced, our yield, from four moles of aluminium reacting with two moles of chlorine.

So the first thing you're going to do is below each reactant in your balanced symbol equation, you want to create a calculation grid, and that grid is going to include the number of moles that are available, the moles that you need.

We're going to then create a ratio between those, so available divided by need in terms of moles, and then we need to identify our limiting reactant.

So the grid might look something a little bit like this for our example.

So once your calculation grid is created, you need to populate it with the number of moles that are available and the moles that are needed.

Now, the moles that are available in this reaction has come straight from our question, so I'm gonna copy those values straight under the appropriate reactant in my calculation grid.

The moles that are needed are the coefficients for those reactants from your balanced symbol equation.

So again, simply copying them into your grid.

Now the next thing you need to do is you need to calculate the ratio of the moles that are available compared to the moles that are needed, and that's one of the reasons why we say available first and then the number of needed because you're just going to divide them.

So again, we're putting them in our grid in the same way that we would put them into our calculator and then we get a value for that.

Once you have your ratio of moles that are available versus the moles that are needed, you can use those to identify your limiting reactant.

All you're going to do is to compare those values and whichever one is smaller is your limiting reactant because that's the lower number of particles that you have available for this particular ratio that's needed for this chemical reaction.

Now, once you've identified which reactant is the limiting reactant, you can move on to then calculate that theoretical yield.

What you're going to do then is take then the limiting reactants available over needed value, so that ratio value, and multiply it by the coefficient of the product you are trying to calculate a theoretical yield for.

And in this situation then, it's going to be 0.

6667 times two, which gives us a value of 1.

333.

Now chemists tend to stick to a standard of three significant figures unless told otherwise for a final answer.

So I would say that for this particular reaction, 1.

33 moles of aluminium chloride can be produced when four moles of the aluminium reacts with two moles of chlorine.

Let's take a moment here for another quick check.

How many moles of carbon dioxide do you think can be produced when three moles of butane burns in eight moles of oxygen? So this may take a little bit of time, so I'm going to recommend that you pause the video and come back when you're ready to check your answer.

Okay, let's see how you got on.

Now for this particular question, I would've hoped that you created that calculation grid that we have outlined previously, and if you've done that, your final answer then should have been letter C.

Now, if you didn't get that answer, I'm going to recommend you pause the video and double check your calculations to see where you may have gone wrong, so you can try to avoid that in the future, but very well done if you manage to get the correct answer.

Great job guys, keep it up.

Now, we know that only by comparing the number of reactant particles that are available, so the number of moles that are available of our reactants, can we actually identify one as being that limiting reactant? The problem we've got is that in a laboratory setting, we don't tend to work by counting out moles.

What we do instead is we measure out a mass or a volume of our reactants.

Now because of that, in those instances, our reactant masses must first be converted into moles using a periodic table.

And that mathematical relationship, that moles is equal to the mass in grammes divided by the relative mass.

But the beauty of this is we can use the similar mathematical relationship in order to calculate the mass of our theoretical yield just by rearranging that a little bit.

And mass in grammes then is equal to the relative mass times the number of moles.

So let's go through an example of how we can look at using values you might collect in the lab about your reactants to then calculate that theoretical yield.

Now, to do this, I'm going to use another example.

I'd like to know the mass of potassium oxide that can be produced when 12 grammes of potassium burns in 25 grammes of oxygen, and I have my balanced symbol equation shown.

So what I'm going to do is very similar to before, I'm going to create a calculation grid here, and you'll notice the bottom four bullet points are exactly the same as what we used before.

All I'm in doing now then is including the mass in grammes and that relative mass because those are the two additional pieces of information I need in order to calculate the number of moles or number of particles that are available in each of those reactant samples that I'm starting with.

Now, once my grid has been formed, I need to populate it and the first thing I need is that mass in grammes, and that's gonna come straight from the question.

So simply copying those up.

Then I need the relative mass for each of those substances, and I'll need to use a periodic table for that and fill that part of my grid in.

Now, I've put the mass in grammes first and then the relative mass because the next thing I need to do is to divide those two to find the number of particles available in that sample that I have.

So simply dividing those values to give me the moles of particles that are available of each.

Now that I have the number of moles that are available, I can complete my grid the same way that I did before.

So I'm going to now populate the next row with the number of moles that are needed.

And remember, those are the coefficients from my balanced symbol equation, and then I'm going to divide then the number of moles that are available by the number of moles that are needed to give me that ratio.

I'm then going to compare those values and whichever value is smaller is my limiting reactant.

Now that I've identified that potassium is my limiting reactant, I can now calculate the number of moles or the number of particles that I'll be able to form of my product, potassium oxide.

So I'll take then the number of moles that are available that I can use and multiply that by the product coefficient, which in this case is two.

And when I get that, I get a value of 0.

1538 moles of potassium oxide could be formed in this situation, but it doesn't stop there.

Remember, I was asked to find the mass of this product that's gonna be formed.

So in order to do that, I now need to find the relative mass for the product I've been asked about.

So using my periodic table, the relative mass for potassium oxide is 94.

Then I can use that relationship of the mass in grammes as equal to the number of moles times the relative mass, multiplying those together and I will get a final mass then of 14.

46 grammes.

But remember using that standard of three significant figures unless told otherwise, we can say that 14.

5 grammes of potassium oxide can be produced when 12 grammes of potassium burns in 25 grammes of oxygen.

Let's go through another example, only this time, rather than writing out each step, I'm going to talk you through it using my calculation grid that's already been drawn here.

I would like to know the mass of ammonia that's produced when I react 15 grammes of nitrogen with 15 grammes of hydrogen.

So I'm going to put those two masses underneath the appropriate reactants, and then I'm going to use my periodic table to fill in the relative mass for each.

Dividing those values then gives me the number of moles or the number of particles that are available in each of those samples.

Next, I need to fill in the number of moles I need for my recipe or my molar ratios for this particular reaction.

And that's the coefficients from my balanced symbol equation.

Again, simply copying those then into the relevant place in the grid.

When I divide then the number of moles I have available by the moles that are needed, I get these values and I can see then that the nitrogen has the lower value and therefore the nitrogen is my limiting reactant.

If I take that value then of 0.

5357 and multiply it by the coefficient of my product, I now know the number of moles or number of particles of ammonia I can produce using these masses of nitrogen and hydrogen.

In order to find the mass of ammonia though, I need its relative mass.

So using my periodic table, I find that it's 17.

If I multiply its relative mass by the number of moles, I get an idea of the mass of ammonia that can be produced using these substances and to three significant figures then, my answer is 18.

2 grammes of ammonia produced.

What I'd like you to do now then is to use a similar method to find the mass of carbon dioxide that can be produced from three grammes of carbon reacting with five grammes of oxygen.

And as before, I'd like you to give your answer to three significant figures.

As a reminder to help you with the processing, I've just indicated them again on the left hand side example, but you'll need to grab a pen, some paper, a periodic table, and a calculator, pause the video and come back when you're ready to check your answer.

Okay, let's see how you got on.

Now, if you've done your calculations correctly, you should have had a final mass of 6.

88 grammes of carbon dioxide that is produced to three significant figures.

If you didn't get that answer, you may wish to pause the video so you can double check your calculations and the values you've put into your calculation grid to see where you may have gone wrong so you can look to avoid that in the future.

But incredibly well done if you managed to get that final answer of 6.

88 grammes.

Great job, guys.

Let's move on to the last task then of today's lesson.

In this first part, we want to consider the reaction of tin with water.

Now, tin normally is unreactive with room temperature water, but if that water is heated up so that it's steam, it will tend to react as shown in the equation below.

What I'd like you to do is to consider each of these different scenarios to determine which reactant is limiting in each case, and then use that information to calculate the number of moles of each product that will form for that scenario.

So pause the video and come back when you're ready to check your answers.

Okay, let's see how you got on.

So for the first one, we had four moles of tin and one mole of water.

And when we go through our calculations, water is our limiting reactant.

So give yourself a mark for that.

Using that information then, we need to then multiply the 0.

5 value we got for water by the coefficient for our products, and we have then 0.

5 moles of tin oxide produced and one mole of hydrogen produced.

So give yourself one mark each for those products.

If we move to letter B then, we had three moles of tin this time and 10 moles of water.

And in this situation it was the tin that's the limiting reactant.

And as a result, we have three moles of tin oxide produced and six moles of hydrogen produced.

Now, if you didn't get that answer, make sure that you are pausing the video to double check your calculations and see if you can identify where you've gone wrong, but well done if you got this correct.

Now, for C, we had 0.

5 moles of tin and 0.

25 moles of water.

And in this case, the water is limiting and that means we have produced 0.

125 moles of tin oxide and 0.

5 moles of hydrogen that's produced.

So well done if you got this correct.

For the next part of this task, I'd like you to consider the reaction of sodium that's used to extract titanium from titanium chloride using the reaction that's shown.

So I'd like you to essentially calculate the theoretical yield of titanium, but I've broken it up into more manageable sections for you for a step by step because this is essentially how you'd get the marks if you are asked a question like this on an exam paper.

So have a go and come back when you are ready to check your answers.

Okay, let's see how you got on.

So for 2A, you were asked to calculate the number of moles that are present in the samples of our reactants, and we had 10.

9 moles of sodium and 2.

63 moles of titanium chloride that were present.

So you'd get one mark for being able to calculate both of those moles.

Well done if you got those correct.

Part 2B then asks you to decide then which of our reactants is the limiting reactant.

And if we carry through our calculation grid that we learned about earlier in the lesson, we can see that the titanium chloride was that limiting reactant, especially if we used our values from part 2A.

It's worth remembering whenever you get a question if it's broken down into parts A, B, C, and D, remember, you can use any information from the previous questions to help you in those subsequent questions.

So don't think that you can only use the information that's provided right there.

You can go back to those previous questions to help you.

So well done if you've got titanium chloride as your limiting reactant.

Now, part 2C then asks you to calculate the theoretical yield of titanium that could be formed from those samples of reactants.

Now, the first thing you need to do is to identify the number of particles of titanium that is forming.

So that's using information from part 2A and 2B, how many moles were able to react and which was your limiting reactant.

Now, when we do that, we find that 2.

63 moles of titanium were going to be produced.

And then once we find the relative mass for titanium, we know then that 126 grammes of a titanium is being produced in this particular reaction.

And we're on to the last part of the last task of today's lesson.

What I'd like you to do is to calculate the theoretical yield of carbon dioxide that's produced when 25 grammes of propanol is burned in 15 grammes of oxygen.

So pause the video and come back when you're ready to check your answers.

Okay, let's see how you got on.

Now, if you've carried out your calculations correctly, you should have got a final answer of 13.

8 grammes of carbon dioxide that's produced.

If you didn't get that answer, definitely pause the video so that you can compare your values against the answer key that's shown.

But as a good guide, it's worth remembering that if you've set up your calculation grid in the same order every time, we should be dividing as we go down the reactant side and multiplying as you go up the product side.

As a general guide.

But very, very well done if you managed to get that final answer correct guys, you just amazing today.

My goodness, we have gone through a lot in today's lesson, and I would not be surprised if you were feeling overwhelmed at times.

So let's just take a step back and summarise what we've managed to go through in this time.

We reminded ourselves that in a chemical reaction, all the reactant atoms rearrange to form all of the products, but when that happens, sometimes there's a reactant that's added to excess.

And what that means is that not all of it has reacted, and because of it, some of these excess reactants will be found in the final product mixture.

By contrast, a limiting reactant is one that has fully reacted.

Every single particle is reacted, but that has restricted or limited the amount of product that can actually form from those available particles.

But there's some use to this.

The moles or the number of particles that we have of our limiting reactant can help us then calculate the theoretical yield or the amount of product that can be formed in a reaction in a particular scenario.

And that that theoretical yield is calculated by using the stoichiometry or the molar ratios that are shown in a balanced symbol equation.

And that mathematical relationships of the number of moles is equal to the mass of our sample divided by the relative mass of the particles we're talking about.

Now, what I wanna do is just take a moment here and recognise that this is not an easy topic, okay? If you were feeling overwhelmed and just feeling very, very worried about this, please be kind to yourself.

This is not something that people get the very first time they go through it.

This is gonna take some practise and a bit of perseverance, but hopefully what you're seeing here is that we're using similar mathematical processes in various ways and what we want to look at then is what comes up most frequently.

That's where we wanna start feeling most comfortable with, and then looking at how we can extend our ability to include those other steps, okay? You are doing so well.

I am incredibly with how you're doing, and what we need to do is just keep up.

This is one of those topics that you just need to come back to every once in a while, remind yourself about it to keep that confidence up, but it will come.

You just need to persevere.

I've had a really, really good time learning with you.

I hope you had a good time learning with me, and I hope to see you again soon, bye for now.