Lesson video

Hello, my name's Mrs. Niven, And today, we'll be looking at how we can determine a chemical equation from given values as part of our unit on calculations evolving masses.

What we go through in today's lesson, you may have some experience of from your previous learning.

But what today's lesson is all about is being able to link up the values that we are able to collect from a practical experiment and how we can use those to communicate information about the reaction that's taken place to other people, and also give us some insight about the particles that are involved in that reaction.

So by the end of today's lesson, you should hopefully feel more comfortable being able to deduce a balanced symbol equation from the masses that you have about the reactance or products for a given reaction.

Now, I'll be referring to some keywords throughout this lesson and they include ratio, stoichiometry, balanced symbol equation, and mole.

Now, the definitions for these keywords are provided in sentence form on the next slide.

And you may wish to pause the video here so you can have a quick read through or maybe jot down a note of what each means, so that you can refer back to it later on in this lesson or later on in your learning.

So, today's lesson is broken into three sections.

First, we'll look at chemical ratios in general before moving on to look at how we can calculate the reacting molar ratios for a reaction, and finally, finish the lesson by looking at how we can use those ratios in order to balance a chemical reaction.

So, let's get started by looking at what we mean by chemical ratios.

Now, whenever we use that term ratio, what we're talking about is relationship between two or more quantities.

So it doesn't matter if you're talking about a ratio in chemistry, in building, or in maths.

What a ratio does is it allows us to determine the size of one quantity in relation to another.

And I'll give you an example.

Whenever somebody makes concrete, standard concrete has a ratio of one part cement, two part sand, four part aggregate, which is like pebbles or stones.

And then, all of those different parts are mixed together with some water simply to bind them.

But it's a standard ratio and that ratio could change for concrete depending on its use later on.

But it's a ratio.

It's telling us the quantity of one material compared to another for that final product.

Now, ratios exist in chemistry in a variety of ways.

For instance, there are ratios of elements within the substances that we're using for different reactions.

And those ratios show up as the subscript numbers in the chemical formula.

For example, here in H2O, those subscript numbers would be a 2 and a 1.

But what do those values actually mean? Well, it means I have two atoms of hydrogen for every one atom of oxygen.

And if I wanted to represent one molecule of water, then I would need to show two hydrogens shown here as the white balls and one oxygen shown here with one red.

And the thing is, if I have multiple molecules of a substance present, the ratio of the atoms within that substance doesn't change.

So if I have multiple molecules of water, the ratio of the atoms within each molecule of water stays the same, two hydrogens for every one oxygen.

And the same is true if we look at larger substances.

So if we're looking at a giant covalent substance or even an ionic substance, the chemical formula for these substances indicate those ratios of atoms or ions in a formula unit rather than the molecule.

So, no matter the size of the structure, the ratio of the atoms or the ions in those substances stays constant.

So if I look at this example of silicon and oxygen that have bonded together in this giant covalent structure and I take a closer look at a part of it, I can see that for every two silicon atoms there are four oxygen atoms bonded around it.

So, I could say that it's formula was Si2O4, but we simplify it down for its chemical formula.

And so, this is SiO2 or silicon dioxide.

If I look at an ionic substance, here, I've got sodium ions bonded to chloride ions, and I click a really close look at the number of ions in these.

I can see that for each formula unit, I have one sodium ion for every one chloride ion.

So its ratio is one-to-one, and so its chemical formula will be NaCl for sodium chloride.

Let's stop here for a quick check.

True or false, the ratio of atoms or ions in a substance changes as the number of particles increases.

Well done if you said false.

But which of these statements best justifies your answer? Well done if you said a, the ratio of atoms and ions in a substance is indicated by its chemical formula.

It does not change if the size of the structure increases.

Well done if you got those correct, guys.

Great start.

So while ratios exist between the elements within the substance, ratios also exist between substances.

And those ratios are indicated by the stoichiometry or that molar ratio that we can see in a chemical equation.

And what that indicates is that regardless of the amount that you're actually using of each substance, that molar ratio for a chemical reaction doesn't change.

So, let's look at an example.

I have a reaction here, and the stoichiometry or that molar ratio between substances has been circled.

So, these are the coefficients for my balanced symbol equation, those large numbers before each chemical formula.

And it tells me regardless of how much I'm starting from with each of my substances, the molar ratio or the relationship between those substances in this chemical reaction.

So, let's say for instance that rather than having two moles of my aluminium, I only have 0.

5.

I can use this molar ratio to find out what that change in molar ratio will be using my balanced symbol equation in my understanding of ratios.

To get from two to 0.

5, I need to divide by four, which means I'd need to divide all the other coefficients by four as well to get that new molar ratio.

Similarly, let's say now rather than two aluminium, I have 10 moles of aluminium.

Well, to get from two to 10, I need to multiply by five and therefore, I need to multiply all the other coefficients by five as well to get that new molar ratio.

And it's all because of that relationship between the substances in this chemical equation.

Let's stop here for another quick check.

Which chemical equation or equations shows a similar molar ratio to the reaction shown? Now, you may wish to pause the video here so you can take a closer look at each of these equations before making your final choice.

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

Okay, let's see how you got on.

Well done if you chose b and d as showing a similar molar ratio to the reaction shown.

B, all you needed to do was multiply every one of those coefficients by two.

And for d, all of those coefficients were divided by two.

A and c do not work, because not all of those coefficients were changed.

Only one or two of them have been.

So, you've gotta be very careful in identifying those molar ratio reactions.

But very well done if you manage to get at least one of those correct and incredibly well done if you manage to find both of those correct answers.

Great job, guys.

Okay, time for our first task.

So, what I'd like you to do to start with is to match each of the chemical formulas that are shown to the correct diagram.

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

Okay, let's see how you got on.

So C6H12O6, I'm looking for something that has at least three different atoms in it, and that's gonna be the third diagram.

Na2O is going to be the last diagram or the one on the far right.

Ki, for me, I have two different atoms in there, but it's a metal and a non-metal, which means I should be looking for an ionic structure, and that's going to be the first diagram.

So, C4H10 then is the second diagram from the left.

Well done if you managed to get those correct.

Great job, guys.

For the next part of this task, I'd like you to use the stoichiometry or those molar ratios to determine the new molar quantities for each of the parts, a through d, shown below the chemical equation.

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

Okay, let's see how you got on.

So, the first thing you needed to do was identify what the stoichiometry was of the original equation.

And then for a, you needed to multiply by two and do that for the rest of those values, so 4, 2, and 12.

For b, you were dividing by two and giving you values of 1.

5, 0.

5, and in 3.

For c, you were dividing by six and getting values of 0.

5, 0.

33 and 0.

17, sorry.

And for d, which was particularly tricky, you needed to multiply by three and divide by four.

So, that's essentially multiplying by three quarters.

And that gives you then the final values of 2.

25, 1.

5, and 4.

5.

Very, very well done if you manage to get those correct, guys.

Great job.

For the next part of this task then what I'd like you to do is to consider the reaction of methane burning in oxygen.

And when it does so, it produces, when it completely combusts, carbon dioxide and water.

So, I'd like you to decide which of the four chemical equations that are shown is correct and also identify what is incorrect about the other chemical equations.

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

Okay, let's see how you got on.

Hopefully, you have chosen b as being the correct chemical equation.

Very well done if you managed to get that.

The other three then you needed to identify what was incorrect about them.

And for a, carbon dioxide had the incorrect ratio.

It should be CO2.

For c, it also had an incorrect ratio.

This time for water, which should be H2O.

And for d then, it had the incorrect stoichiometry.

So, the molar ratios were incorrect.

It should have been 2O2.

Well done if you managed to get those correct, guys.

What a fantastic start to today's lesson.

Keep it up.

Now, that we're feeling a little more confident talking about chemical ratios in general, let's look at how we can calculate then reacting molar ratios.

Now, we said earlier that the stoichiometry of a reaction is shown by a balanced symbol equation and that the coefficients in that symbol equation represents the molar ratio of the substances involved in it.

Now, we need to remember that one mole is equal to or represents 6.

02 times 10 to the 23 particles of any substance.

So if we look at this particular reaction of creating ammonia, we could read it as one mole of nitrogen molecules reacts with three moles of hydrogen molecules to form two moles of ammonia molecules.

Now, the thing is sometimes we don't actually know what the reacting ratio of some chemicals are at the start, but we can at least use the mass of the reactants that were used to calculate that reacting molar ratio.

And these calculations then are going to use both that mathematical relationship we've been using throughout the entire topic of massing grammes is equal to the relative mass times moles and also our understanding of ratios.

Now, in order to do these calculations, you will need a calculator, a periodic table.

You'll need the masses for the reactants in those reactions, and you'll also need a step-by-step strategy.

Now, you'll be provided the masses and the strategy in a moment, but if you don't have a calculator or periodic table to hand, I reckon you should pause the video and come back when you have those before we continue.

So, you have your periodic table.

You have a calculator.

You'll be provided the masses of your reactants.

And what I'm going to do now then is use an example to talk you through a strategy that you could use in order to calculate those reacting molar ratios.

So, I want to know what the reacting ratio is when 0.

892 grammes of sulphur dioxide reacts with two, sorry, 0.

224 grammes of oxygen to form this sulphur trioxide.

So, the first thing I'm going to do is I want to write down the chemical formula for each reactant.

So I find my reactants, write down the names below them.

I'm going to put the formulas.

So, sulphur dioxide is SO2 and oxygen is O2.

It's the formula for the element, not its symbol.

So it's important that oxygen is O2 and not just O because it is a diatomic element.

Then below those reactants, I'm going to create a calculation grid that includes some important information that I'm going to need for these calculations.

And that's going to be the mass in grammes, the relative masses, the number of moles within that mass in grammes, the unitary ratio, and then finally, its molar ratio.

So, you should have something that looks a little bit like this once your calculation grid has been created.

Once you have your calculation grid sorted, you need to populate with some information.

And the easiest thing to do is the masses from your question for each of your reactants.

You're simply going to copy those into your calculation grid.

The next thing you can do then is to use your periodic table to calculate those relative masses for each of those reactants and add those to your grid.

The next step then is for you to use that mathematical relationship of moles as equal to the mass and grammes divided by the relative mass in order to calculate the number of moles.

So, we're looking at finding the number of particles that are in each of those mass samples.

So, there's a reason this calculation grid is set up the way.

It is because it's the massing grammes divided by that relative mass.

So, you're literally adding the numbers in order from top to bottom how they go into your calculator.

And then, the moles row is for what is coming outta your calculator.

So, you can record those going forward.

The next step then is for you to calculate the unitary ratio of your reactants.

And the easiest way to do that is to identify the lowest value for your moles row and divide all the values by that lowest moles value.

So for this example, it's the oxygen 0.

007.

And when I divide all the values by the 0.

007, I get 1.

989 for sulphur dioxide and a 1 for oxygen, and that's really important.

What we should always find when you are calculating a unitary ratio is that at least one value should come out as a value of one.

If not, you've gone wrong somewhere.

Go back and double check your work.

Finally, then you are going to calculate that molar ratio.

Now, the thing to remember is that molar ratios should be whole numbers.

And some unitary ratios, if they're not a whole number, could be easily rounded to achieve a whole number.

And that's exactly what happens in this example.

That sulphur dioxide value of 1.

989 can easily be rounded to a value of two.

And because of that then, I get a final answer of sulphur dioxide to oxygen from my reacting masses will have a ratio of two to one.

Okay, let's go through another example, but this time I'm not going to show you each of these steps.

I'm going to talk you through them.

So, the first thing I need to do is to identify my reactants and put the formula there, and then I need to outline my calculation grid below that.

I'll then populate my grid with the masses from my question and the relative masses for my substances using my periodic table.

The first calculation I need to do then within my grid is to calculate the number of moles.

So, dividing the mass by that relative mass.

To find the unitary ratio, I first need to identify the lowest moles value that I'm going to divide all of the other values by.

And when I do that, my magnesium chloride comes out as 1, and the sodium hydroxide comes out as 2.

003.

To find my molar ratio then, I'm looking for whole numbers and I need to round the sodium hydroxide to a whole number and it comes out as two, which gives me a final answer of magnesium chloride to sodium hydroxide is going to be a one to two ratio.

What I'd like you to do now then is to have a go at your own calculations.

I'd like you to find the reacting ratio for when propane reacts with that mass of oxygen.

And as a guide, I'm showing you the calculation processing that I used on my grid again on the left hand side.

This is gonna take a little bit of time, so grab your calculator, that periodic table, and a pen.

Pause the video and come back when you're ready to check your answer.

Okay, let's see how you got on.

So if you've done all your calculations correctly, you should have had a one to five ratio for your propane to oxygen.

Now, if you didn't get that answer, I recommend that you pause the video and double check how your calculations compare to the answer key that has been shown.

But incredibly well done if you manage to get the correct reacting ratios, guys.

Great job.

Now, it's important to understand that sometimes that unitary ratio value is not close enough to whole numbers that it could actually be easily rounded.

And in those situations, what you need to do is multiply accordingly in order to achieve a whole number, in order to remove the decimal.

And I always found it useful to have a little bit of a guide, and that's what this table represents.

If you have a decimal that ends in a roundabout 0.

5, well, if we represented that decimal as a fraction, it would come up as about one-half.

And I know that if I wanna get a whole number from a half, I need to multiply by two.

The same goes for the other decimals then.

Roundabout 0.

33 is about one-third, so I'd have to multiply by three.

So if you can change your decimal to a fraction, the denominator of your fraction tells you what you need to multiply by in order to get a whole number ratio.

So, let's look at an example.

If I have a unitary ratio of 1.

52 to one, I can see that that 0.

52 is very close to around about a half.

So, I need to multiply by two.

But the key here is that when you're multiplying, you are not just multiplying that one value.

You need to multiply all of those unitary ratio values by two.

And when we do that, we get values that are closer to whole numbers.

So this time, 3.

04 can very easily be rounded now to three.

And because I had to multiply all of my values by that value of two to remove the decimal, my overall final molar ratio would be a three to two.

Let's stop here for a quick check.

Which of these ratios do you think need to be multiplied accordingly in order to achieve a whole number ratio? And if you wanna challenge yourself a bit, what number would you multiply those by in order to achieve that whole number ratio? Well done if you chose a and d.

Both of those have decimals that cannot be easily rounded to whole numbers, so they'll need to be multiplied accordingly.

For a, you would've needed to multiply by two because not 0.

45 is very close to not 0.

5.

And d, you would've needed to multiply by four because 0.

77 is very close to 0.

75.

Very, very well done if you manage to identify the two ratios that needed multiplying and incredibly well done if you managed to choose the correct value to multiply those by.

Great job, guys.

Let's try another example.

I want to know the reacting ratio for when 1.

45 grammes of aluminium reacts with 20.

5 grammes of iodine.

So, the first thing I need to do is identify my reactants and then write the chemical formula for each.

And below that, I'm going to create my calculation grid and populate it with the mass for each of those substances.

Then, I grab my periodic table and calculate the relative masses for each of those substances and divide them then to find the number of moles or the number of particles that are present in each of those mass samples for my reactants.

When I then find the unitary ratio, I'm dividing by that lowest mole value.

And when I do so, I can see that the iodine unitary ratio value does not come out as a whole number or one that can be easily rounded.

So, I'm gonna need to multiply accordingly to get a whole number.

And for this, I'm going to multiply by two.

And when I do so, I get values that are far more easily rounded to a whole number.

So, my final answer for my reacting masses for this question is aluminium to iodine is going to be two to three ratio.

What I'd like you to do now then is to calculate the reacting ratio when 8.

7 grammes of gallium reacts with 3.

0 grammes of oxygen.

And to remind you of the processing, I've put them back up again on the left hand example that I've just gone through.

You are going to need now your periodic table and a calculator and a little bit of time.

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

Well done if you managed to get a gallium to oxygen ratio of four to three.

Very, very well done.

Now, if you didn't get that value, definitely pause the video so that you can go back through your calculations and double check where you've gone wrong in the hopes that you can try to avoid those going forward.

But very, very well done if you managed to get that answer, guys.

Great job.

Okay, time for the next task in today's lesson.

What I'd like you to do is to use the data that's been provided to calculate the reacting molar ratio in each instance.

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

Okay, let's see how you're going on.

So for a, you should have had a final reacting ratio for the barium to chlorine of one to two.

If you didn't get that answer, please do make sure that you're pausing the video to double check your calculations to see where you may have gone wrong, so you can avoid them going forward.

But very well done if you've got that correct answer, guys.

For b then, you should have had a final reacting ratio of potassium hydroxide to sulfuric acid of two to one.

Very well done if you got that.

And for c, then you should have got a final reacting ratio of magnesium oxide to phosphoric acid of a three to two ratio.

This one mean you had to multiply that unitary ratio to get whole numbers.

So, very well done if you managed to get these.

I'm really impressed by your perseverance today.

Keep it up.

Now, that we're feeling more comfortable calculating the reacting molar ratios, let's look at how we can use these to balance a chemical equation.

Now, we said earlier that the reacting ratios that we did previously indicate then the relative amounts of each reactant that's required for a reaction to occur.

And for this one then if we looked at our example from earlier, sulphur dioxide to oxygen, we calculated as being a two to one ratio.

Now, that essentially means that I'm going to need twice as much sulphur dioxide to react with oxygen in order to make my product.

Now, there are two ways that I can use this information in order to write a balanced equation for that reaction.

One is to simply use these as my coefficients and then write a balanced equation.

The other is a little bit more complicated, but still quite easy.

If I know the mass of my products as well as the mass for my reactants, I could actually calculate the stoichiometry or those molar ratios for my products, as well as my reactants, in the same way that I calculated those reacting ratios in the previous part of this lesson.

And that then allows a chemist to calculate the stoichiometry for the entire reaction rather than just the reactants.

And then, I can use the values from those calculations to write a balanced symbol equation from those known masses.

So overall, once I know that reacting ratios, I can write a balanced symbol equation.

So again, like we said, I'm gonna need twice as much sulphur dioxide to react with oxygen to make a product.

And if you recall, that product was sulphur trioxide.

Now, I told you earlier that there were two ways that we could use this information to write a balanced symbol equation.

And the first one was simply to balance that equation.

So, I'm gonna go through that process first.

The first thing you want to do is write the chemical formula, including any of the reacting ratios that have for your reactants and your products.

Now, the reactants for this one is particularly easy 'cause we already had our sulphur dioxide and oxygen, and we already knew that that reacting ratio was two to one.

Sulphur trioxide is a formula you could figure out is SO3, but we don't know the reacting ratio here.

And we're gonna need to figure that out.

And we do that simply by balancing the equation the way we do simply understanding the idea of conservation of atoms. So, I'm going to separate my reactants from my products.

I'm going to write, draw a little diagram to represent each of my substances.

So, I've got my molecules, the sulphur dioxide, and I've drawn two because I have a 2 for my coefficient, one of my oxygen molecules, and then sulphur trioxide as a molecule as well.

And then, keeping track of the amount of sulphur and oxygen on my reactants in products.

So for my reactants, I have two sulphur and six oxygen atoms. And on the product side, I have one sulphur and three oxygen atoms. Now these are clearly unbalanced, and in order to balance these numbers, I need to another molecule of sulphur trioxide.

And when I add that diagram to my picture here and then recalculate the number of sulphur and oxygen atoms, I can see it's now balanced.

So, I count up the number of diagrams I have for my sulphur trioxide, which is two, and I have now my final balanced symbol equation of 2SO2 plus O2 makes 2SO3.

Now, the issue with using reacting ratios in order to write a balanced symbol equation, if you just stay with your reacting ratios and then you have to swap your head round to how do we balance an equation.

Again, using this conservation of atoms using two different processes.

And sometimes, it's not as easy to swap your head round.

If you're given the mass for the products, you can actually use the exact same process you use to find the reacting ratios, but this time, you're finding the reacting molar ratios for every substance in your reaction.

And you can use that to find your balanced symbol equations.

So, it's kind of saving yourself a little bit of brain space.

I'm gonna walk through how to do that now, but just be aware that at the end of this, it doesn't matter which process you use as long as you're comfortable with it.

But if you're asked to find use molar ratios to determine a balanced symbol equation, this is the process that I would recommend.

So same as before, you're going to identify the substances in your reaction and write down that chemical formula below it.

Below that then, you're going to create your calculation grid and populate it with the massing grammes from the question.

And using your periodic table, fill in the relative masses for each of the substances.

Divide those values to find the number of moles or the number of particles that are present in that mass sample.

And then using the smallest mole value, divide them all by that, so here 0.

007, to find your unitary ratio.

At this point then, you need to double check can those values be easily rounded to whole numbers or do I need to multiply accordingly to create whole numbers.

In this example, I can round them all and I now have a ratio of two to one to two.

So, once you have your molar ratio for both your reactants in your products, you just need to figure out how to write that balanced symbol equation, and it's actually a lot easier than you might think.

You identify your reactants, use the formula from above your calculation grid, and simply write that down.

Same thing then for your products.

Just draw your arrow and write the formula for your product.

Then, all you need to do is go to your molar ratio row and input the values that you've calculated in front of the relevant formula to give you that final answer of 2SO2 plus O2 makes 2SO3.

And voila, balanced symbol equation from molar ratios.

Okay, let's go through another example.

So same as before, I'm going to identify my substances and write down their chemical formula.

Being careful to remember that chlorine is diatomic and therefore, should be Cl2.

Create my calculation grid and populate it with the mass from the question and the relative masses using a periodic table.

Divide those values to find the number of moles, the number of particles, and each of those mass samples.

And then, identify the smallest value of those moles that you've calculated and divide all of the values in that row by the smallest value, here is 1.

971, to create your unitary ratio.

And when I look at all those values, the iron chloride could be easily rounded, but the chlorine value cannot because it's 1.

529.

And when I look at that a little bit more closely, it's nearly one-half.

So, I multiply then that value and the other unitary ratio values by two, and I get values that are a lot more easily rounded to whole numbers.

Giving me a final answer then of 2Fe plus 3Cl2 makes 2FeCl3.

What I'd like you to do now then is to calculate the molar ratios to determine the balanced equation for a reaction when hydrogen peroxide then decomposes to make water and oxygen.

So, this is going to take a little bit of time.

Definitely pause the video and come back when you're ready to check your answer.

Okay, let's see how you got on.

If you've done your calculations correctly, you should have had a final answer of 2H2O2 makes 2H2O plus O2.

If you didn't get that value, or sorry, that final chemical equation, definitely pause the video and check your calculations to see where you may have gone wrong.

But really, really impressed with how you guys are getting on today.

Just amazing.

Time for the last task of today's lesson.

Now, you may or may not know, but a lot of the metals that we use in today's society are actually needing to be isolated from a compound using some chemical processing.

So, what I'd like you to do is use molar ratios to determine the balanced symbol equation for each of the reactions that are described to isolate a metal.

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 letter a, you were asked to determine the balanced symbol equation for the decomposition of aluminium oxide.

And if you've done your calculations correctly, you should have had a final answer of 2Al2O3 makes 4Al plus 3O2.

If you didn't get that answer, definitely pause the video so you can check your calculations and see where you may have gone wrong.

But well done if you got that correct.

And for part b, then it was a little bit trickier.

However, if you did the calculations correctly and then were able to correctly identify the reactants in products, you should have had a final answer of Fe2O3 plus 3CO makes 2Fe plus 3CO2.

So, incredibly well done if you manage to get that correct as well.

I'm so impressed with the way you guys are working.

Keep it up.

Okay, for the last part of this task, I'd like you to help Alex.

He's used masses to calculate the molar ratio for reaction.

That's using magnesium to extract titanium for titanium chloride.

And when he is double checked his final equation, he's found that the conservation of atoms has not happened.

He's made an error somewhere.

So, these are the calculations that he's done.

I'd like you to identify and correct Alex's error.

And by doing so right now, then the correct balance equation he should have got had he not made that error.

So, you may wish to discuss your ideas with the people nearest you.

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

Okay, let's see how you got on.

So, the first thing I'm gonna do is look at Alex's final answer and compare those atoms, where has he gone wrong.

And I can see that the number of magnesium and the number of chlorine atoms don't match up.

So, I'm gonna look back at his calculations for the only substance that has both chlorine and magnesium in it, and that's the magnesium chloride.

And when I take a closer look at that, he has actually written the incorrect formula for magnesium chloride.

It should have been MgCl2, not MgCl.

And there's actually a knock on effect for having written the incorrect formula here.

The relative mass for that substance is now incorrect because he's used the wrong formula.

It should have been 95.

3, 59.

8.

So, that one error of writing the incorrect formula has had a knock on effect to his calculations.

So, once we fix the errors that Alex has made and redo the calculations, we can find a final correctly balanced equation for this reaction as being TiCl4 plus 2Mg makes Ti plus 2MgCl2.

So, incredibly well done if you manage to identify the correct balanced equation for this reaction, guys.

Great job.

Wow, we have gone through quite a lot of work in today's lesson, but let's just take a moment to summarise what we've learned.

We've learned that there are ratios throughout chemistry.

In one way, we can see them is within substances shown by the subscript values of a chemical formula indicating that ratio of atoms or ions in a molecule or a formula unit.

So, looking at a simple structure or even some of these larger structures, we can still see those ratios.

Ratios also exist in chemical reactions.

So, the stoichiometry or that molar ratio of each substance is shown in those balanced symbol equations using coefficients.

So, there's larger numbers that are shown before the chemical formula in a reaction equation.

And that those stoichiometric values or those coefficients can actually be calculated if we have a known mass or ones that are collected from an experiment using our understanding of ratios.

And that mathematical relationship of the mass and grammes is equal to the relative mass times the number of moles.

So, you guys have just been outstanding today.

I'm really, really impressed with how you've got on.

I hope you had a good time learning with me.

I had a great time learning with you, and I hope to see you again soon.

Bye for now.