- [Voiceover] We know what a chemical equation is,
and we've learned how to balance it.
Now we're ready to learn about stoichiometry.
This is an ultra-fancy word that
often makes people think it's difficult, but it really is
just the study or the calculation of the relationships
between the different molecules in a reaction.
This is the actual definition that Wikipedia gives.
"Stoichiometry is the calculation of quantitative,
"or measurable, relationships of the reactants
"and the products," and you're gonna see in chemistry
sometimes people use the word "reagents".
For most of really our purposes, you can use the word
reagents and reactants interchangeably.
They're both the reactants in a reaction.
The reagents are sometimes for special types of reactions
where you wanna throw a reagent in and see
if something happens, and see if your belief
about that substance is true or things like that.
But for our purposes a reagent and a reactant
is the same thing.
So it's a relationship, the reactants and products
in a balanced chemical equation.
So if we're given an unbalanced one, we know how to get
to the balanced point, balanced chemical equation.
So let's do some stoichiometry.
What I'm gonna do is I'm gonna start, just so
we get practice balancing equations, I'm always
gonna start with an unbalanced equation.
Let's say we have iron(III) oxide.
That's F...
You have two iron atoms with three oxygen atoms.
Plus aluminum, A-L,
and it yields
A-L two O three
plus iron.
So, remember, when we're doing stoichiometry,
first of all, we want to deal with balanced equations.
A lot of stoichiometry problems will give you
a balanced equation, but I think it's good practice
to actually balance the equations ourselves.
Let's try to balance this one.
We have two iron atoms here, right, in this iron(III) oxide.
How many iron atoms do we have on the right hand side?
We only have one, so let's multiply this.
Let's multiply this by two, right here.
All right.
Oxygen, we have three on this side,
we have three oxygens on that side, that looks good.
Aluminum on the left hand side, we only have
one aluminum atom.
On the right hand side, we have two aluminum atoms,
so we have to put a two here.
And we have balanced this equation,
so now we're ready to do some stoichiometry.
So stoichiometry is essentially, look, if I give you,
there's not just one type of stoichiometry problem,
but they're all along the lines of, hey, if I give you
x grams of this, how many grams of aluminum do I
need to make this reaction happen?
Or if I give you, you know, y grams of this molecule
and z grams of this molecule, which one's
gonna run out first?
That's all stoichiometry, and we'll actually do
those exact two types of problems in this video.
So let's say that we were given,
let's say we were given, 85 grams
we were given 85 grams of the iron(III) oxide.
So we're given 85 grams.
My question to you is how many grams
of aluminum do we need?
How many grams of aluminum.
Well, you look at the equation,
you immediately see the mole ratio.
For every mole of this,
so for every one atom we use of iron(III) oxide,
for every one of that, we need two aluminums, right?
So what we need to do is figure out how many moles
of this molecule there are in 85 grams,
and then we need to have twice as many moles
of aluminum, right, because for every mole
of the iron(III) oxide, we have two moles of aluminum,
and we're just looking at the coefficients.
We're just looking at the numbers.
One molecule of iron(III) oxide combines
with two molecules of aluminum
to make this reaction happen.
So, let's first figure out how many moles 85 grams are.
What's the atomic mass, or the mass number,
of this entire molecule?
Let me do it down here.
So we have two irons
and three oxygens.
Let me go down and figure out
the atomic masses of iron and oxygen.
Iron is right here, 55.85.
I dunno, I think it's fair enough to round to 56.
Let's say we're dealing with the version of iron,
the isotope of iron that has 30 neutrons.
It has a mass number of 56, so iron has 56
atomic mass number, and then oxygen we already know
is 16, so iron was 56.
So we know that
this mass is going to be two times 56 plus,
plus three times 16, we can do that in our head.
So one molecule of iron(III) oxide
is going to have, is going to be
160 atomic mass units.
One mole, one mole, one mole,
or 6.02 times 10 to the 23,
molecules of iron oxide...
One of those pop-ups...
Iron(III) oxide, is going to have a mass of
160
160 grams.
In our reaction, we said we're starting off
with 85 grams of iron oxide, how many moles is that?
Well, 85 grams,
85 grams of iron(III) oxide
is equal to 85 over 160 moles,
is .53 moles.
So everything we've done so far in this green
and light blue, we figured out how many moles
85 grams of iron(III) oxide is.
We said for every mole, and we figured out
it's .53 moles, 'cause a full mole would have been
160 grams, but we only have 85.
It's .53 moles.
We know from this balanced equation that
for every mole of iron(III) oxide we have
we need to have two moles of aluminum.
If we have .53 moles of the iron molecule,
iron(III) oxide, then we're going to need twice
as many aluminums, we're going to need 1.06
moles of aluminum, right?
I just took .53 times two.
'Cause the ratio is one to two, for every
molecule of this we need two molecules of that.
For every mole of this, we need two moles of this.
If we have .53 moles, you multiply that by two,
and you have 1.06 moles of aluminum.
All right, so we just have to figure out how many
grams is a mole of aluminum, and then multiply that
times 1.06 and we're done!
So aluminum, or aluminium, as some
of our friends across the pond might say.
Aluminium, I enjoy that more.
Aluminium has an atomic,
well, the atomic weight or the weighted average is 26.98,
but let's just say the aluminium that we're
dealing with has a mass of 27 atomic mass units.
One aluminium is
27 atomic mass units,
so one mole of aluminium is going to be
27 grams,
or 6.02 times 10 to the 23 aluminium atoms
is going to be 27 grams.
So if we need 1.06 moles, how many is that going to be?
1.06 moles
of aluminium is equal to
1.06 times 27 grams,
and what is that, what is that?
1.06
times 27
equals 28.62.
So we need 28.62
grams of aluminium.
I won't write the whole thing there, in order to make,
to essentially use up our 85 grams of the iron(III) oxide.
If we had more than, what was the number,
more than 28.62 grams of aluminium, then there'll
be left over after this reaction happens,
assuming we keep mixing it nice and the whole
reaction happens all the way and we'll talk
more about that in the future.
If we have, and in that situation where we have more than
28.63 grams of aluminium, then this molecule
would be the limiting reagent, because we had more
than enough of this, so this is
what's going to limit the amount
of this process from happening.
If we have less than 28.63 grams of, I'll start saying,
aluminum, then the aluminum will be the limiting
reagent, because then we wouldn't be able to use
all of the 85 grams of our iron molecule
our iron(III) oxide molecule.
Anyway, I don't want to confuse you in the end
with that limiting reagent.
So the next video we'll do a whole problem
devoted to limiting reagents.