Hello hello Melissa Maribel here with
some examples on stoichiometry.
Before we begin, remember there are three different
types of conversion factors for stoichiometry.
The first type is molar
mass. We use molar mass whenever we're
converting from grams to moles or moles
to grams. The second one is a mole ratio
which is found on your balanced equation.
You use a mole to mole ratio whenever you are changing your compound.
And the last
one is Avogadro's number.
You use Avogadro's number whenever you see the
keywords of atoms, molecules, particles, and formula units.
All right, now that we
know that, let's jump right in.
What mass of hydrogen peroxide must
decompose to produce 48.64 grams of water?
And we're given our
balanced equation.
Hydrogen peroxide is that H2O2.
And let's figure out what
we're given as a whole.
So what we're given is the 48.64 grams of water and we're
asked to find our mass or our grams of
hydrogen peroxide.
And as I mentioned hydrogen peroxide is H2O2. So we will need this balanced equation because
we're trying to get from grams of H2O to
a completely different compound of grams
of H2O2. Our plan is to go from grams
of water to grams of H2O2. Now we can't
just instantly go to grams to grams. We
have to go from grams to moles and then
moles to grams. So our first step is to
convert our grams of water to moles of
water using our molar mass. Whenever we go from grams to moles we use the molar
mass of that compound. Then once we have our moles of H2O we will change our
compound entirely using our mole to mole
ratio found on our balanced equation. So
we'll go from our moles of water to
moles of hydrogen peroxide. Now that we
have our moles of hydrogen peroxide we
can then convert that down to our grams
of hydrogen peroxide. Once again using
our molar mass of that compound. So let's set this up.
You want to put your given
on top, that's what you always start with
is your given. So we have our 48.68 grams of water and we're
going to use that first conversion
factor as our molar mass of water and
it's going to be 18.02
grams. You align these two across from
each other because then we will then
cancel out our grams of water.
We now are left with moles of water
and we're going to go back to our
balanced equation to do a mole to mole ratio.
So looking back at our balanced equation
we see that we have two moles of
hydrogen peroxide, so that's why I placed
a 2 here, and for every two moles of
water. That's also why put the 2 moles
on the bottom so they can then cancel
those moles of water would then cancel
and we're at moles of hydrogen peroxide.
Now from moles of hydrogen peroxide we
will then use our last conversion factor,
the molar mass of hydrogen peroxide to
get two grams of hydrogen peroxide
because our moles would then cancel and
we would be left with just the grams of
hydrogen peroxide which is what we're
solving for. So multiply straight across,
divided by that 18.02
times the 2 and you get 91.83 grams of hydrogen peroxide.
Our next example states, how many
molecules of carbon dioxide are required
to react with a hundred and seventy four
grams of carbon monoxide?
So remember that carbon dioxide, that "di" meaning
2 is our CO2.
carbon monoxide, "mono" meaning 1, is our CO.
So let's identify what we're given and what we're solving for.
So we're given that 174
grams of carbon monoxide or CO and we
are finding molecules of CO2, key word
right there, molecules.
So remember anytime you see that word
molecules you use Avogadro's number.
Also, we're changing our compound so we have
to use a mole to mole ratio using this
balanced equation.
Let's set it up.
Setting up our plan we see that we're going from grams
of CO or
carbon monoxide and we're trying to get
to molecules of carbon dioxide. So we're
changing our compound, we have to convert
first to moles of carbon monoxide. So
from going from grams to moles we use
the molar mass of carbon monoxide, that's
our first conversion factor. Then, now
that we're at moles of carbon monoxide,
we will change our compound to the moles
of carbon dioxide using our balanced
equation using that mole to mole ratio.
Now that we're at moles of carbon dioxide we
can then convert this to molecules of
carbon dioxide using Avogadro's number.
Your proper set up would then begin with
our grams of carbon monoxide, because
that was our given, and then aligning
those grams across from each other so
they can then cancel. We would then get
moles of our carbon monoxide. Next we're
right here on our second conversion
factor where we have to use our balanced
equation. So these numbers, if we go back
to our balanced equation, we would then
see that that 2 moles of carbon
dioxide came right here from the
coefficient from our balanced equation.
This other 2 came from right here the
other 2 on our balanced equation. That's
where we get our mole to mole ratio from.
Align those moles across from each other,
specifically the moles of carbon
monoxide, and those would cancel.
Now we're at moles of CO2 and then we want
to cancel out our moles of CO2 to get
to molecules of CO2.
So using Avogadro's number on top, we'll
see that the moles of carbon dioxide
would cancel and we would be left with
molecules of CO2. Make sure to multiply
straight across and divide by, these two multiplied,
and you get 3.74 x 10^24 molecules of CO2.
Now it's okay to 
need some more help and to need to see
more practice problems or examples.
That's actually exactly why I've done so
many videos on this so check them out,
I've gone into even more and more detail
with more and more examples to really
understand this topic because you will
keep seeing stoichiometry in
chemistry
all throughout. Now if there's a specific
type of question that you have on
stoichiometry, feel free to leave a comment
below and I'll answer it. And make sure
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time.
