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## Limiting reagent stoichiometry

Current time:0:00Total duration:19:20

# Limiting reactant example problem 1

## Video transcript

We're told methanol, which is
used as a fuel in racing cars and fuel cells, can be made
by the reaction of carbon monoxide and hydrogen. So this is the methanol
right there. They're giving us 356 grams
of carbon monoxide. So carbon monoxide we have
356 grams of it. And they're giving us 65
grams of hydrogen, of molecular hydrogen. They're mixed and allowed
to react. They say what mass of methanol
can be produced? And then they say what mass of
the excess reactant remains after the limiting reactant
has been consumed? So that tells you this is a
limiting reactant problem, that we have too much
or too little of one of these two reactants. These are the two
reactants there. The one that we have less of is
the limiting reactant and that'll dictate how much of the
product we can produce. And the one that we have more
of is the excess reactant. But first we have to figure out
which is the limiting and which is the excess. And before we even do that, we
should always check that our equation is actually balanced. So let's just check that. On the left-hand side of this
equation we have 1 carbon right there. On the right-hand side we also
have 1 carbon, so that looks good so far. On the left-hand side of the
equation we have 1 oxygen, and on the right-hand side
we have 1 oxygen. Looks good so far. Left-hand side we have 4
hydrogens, 2 times 2. On the right-hand side we have
4 hydrogens, 3 plus 1. So this is balanced so we can
proceed to try to figure out what the limiting reactant is. So the way we want to do it is
to figure out how many moles of each of these we're given,
and then figure out how many moles the stoichiometric
ratio that's required by this reaction. We already know what it is. We know that for every two moles
of hydrogen required, we require-- you can see it right
here from the equation-- 1 mole of carbon monoxide. This is what the equation tells
us, so let's see how many moles of hydrogen and how
many moles of carbon monoxide we have. The first place to start--
we've done this several times-- is what is
the molecular weight of carbon monoxide? So carbon monoxide molecular
weight-- let me write it here-- molecular weight
for carbon monoxide. It's a carbon, has a molecular
weight of 12-- it's good to memorize that, carbons and
oxygens and hydrogens show up so frequently. And then oxygen is 16. So it's equal to 28-- a
molecular weight of 28, which tells us that carbon monoxide,
if we want to view it this way-- so let's do the carbon
monoxide first. So we're dealing with the carbon
monoxide. So we have 356 grams of carbon
monoxide, and we want to write this in terms of how many moles
of carbon monoxide do we have. So what we want is how
many grams per mole-- but I actually write-- 1 mole
has how many grams? So 1 mole of carbon monoxide
has how many grams? Well it's molecular weight
is 28 so if we have a mole of them. If we have that 6.02 times 10
to the 23rd molecules of carbon monoxide. That's going to weigh 28 grams.
So 1 mole of carbon monoxide is 28 grams, or I
guess you have 1 mole per every 28 grams. The reason why
I wanted to put the grams in the denominator so it cancels
out over here. So those cancel out
when we multiply. And so we are left with 356
times 1 divided by 28 moles of carbon monoxide. So let's figure what that is. That's 356 divided by 28--
right, that's essentially what we're doing-- divided by
28 is equal to 12.71. Let's just say it's 12.7 moles
of carbon monoxide. Now let's do the same thing
for the hydrogen. We have 65 grams of molecular
hydrogen. What's the molecular
weight of hydrogen? I'll do this in green
over here. Molecular weight of hydrogen
is-- well each hydrogen atom has a molecular weight of 1
times 2, which is equal to 2. So we have 65 grams of a
molecular hydrogen, and the same way, we want to
write it in moles. So we're going to multiply it
times-- 1 mole of hydrogen is equal to how many grams? Well we just figured it out. 1 mole is equal to its molecular
weight is 2. So a mole of it is going to have
a mass of 2 grams. You can view this as 2 grams per
mole or 1 mole per 2 grams. And we want the grams in
the denominator so it cancels out over here. So let's do the math. That cancels out with that. And we have 65 times
1 divided by 2. 65 divided by 2 is what--
32.5 moles of hydrogen. Now we know exactly how many
moles of carbon monoxide and how many moles of hydrogen
they've given us. Let's figure out what
the ratio is. And we'll do it in the exact
same way as we wrote up here. So we have 32.5-- this is what
we're given-- moles of hydrogen-- let me write
that a little bit neater-- moles of hydrogen. And we're given 12.7 moles--
I'll do that in the same color-- 12.7 moles of
carbon monoxide. So if we were to just divide
this, what is this? I guess you could imagine,
divide the numerator and the denominator by 12.7, or just
divide 32.5 divided by 12.7, what do we get? 32.5 divided by 12.7--
no, that's not-- let me write that again. 32.5-- man, these buttons are--
let me do it with the keyboard-- 32.5 divided by
12.7-- that's a lot better-- is 2.5-- I'll just say 2.56. So this can be re-written as--
so if I just re-write this-- I should have written it here to
begin with-- I could write this as we have 2.56 moles of
molecular hydrogen for every 1 mole of carbon monoxide. So this is what we need for
a reaction to occur. This is what the balanced
equation tells us. It tells us that we need 2 moles
of hydrogen for every mole of carbon dioxide. Based on what they've given us,
we just figured out that we have 2.56 moles of
hydrogen for every mole of carbon dioxide. So we have more than
enough hydrogen. We only need 2 for every
mole of carbon dioxide. We have 2.56. So we have an excessive
amount of hydrogen. So the excess reactant
is the hydrogen. Hydrogen is excess reactant. And the other one's going to
be the limiting reactant. We don't have enough carbon
monoxide to react all of the hydrogen, right? We only have 1 for every 2.56. We would actually, for this,
says you need 1.25 or whatever for every, or 1.28
for every 2.56. It should be a 1:2 ratio. So we don't have enough
carbon monoxide to react all of the hydrogen. So carbon monoxide is the
limiting reactant. Now given that this is the
excess reactant, we can use the stoichiometric ratios to
figure out how much methanol's going to be produced. It's all going to be limited
by our carbon monoxide. Did I just say-- hydrogen's
not the limiting reactant, carbon monoxide is the
limiting reactant. We have more than
enough hydrogen. So how much carbon monoxide
do we have? We already figured it out. We have 12.7 moles of
carbon monoxide. And we look at our-- let me
write this over here-- so we have 12.7 moles of
carbon monoxide. And looking at our original
equation, we see for every mole of carbon monoxide, we
produce 1 mole of methanol. So let's write that down. Times-- and we want the carbon
monoxide in the denominator. So for every 1 mole of carbon
monoxide used we have 1 mole of methanol, which is, what,
CH3OH-- did I get that right, yup-- produced. We get that straight from
the balanced equation. And this math is pretty
easy, but it gives us the right unit. Remember, we're using the
carbon monoxide, not the hydrogen because the
carbon monoxide's the limiting reagent. That's what's telling us what's
going-- if we used hydrogen as the limiting
reactant, then we wouldn't have enough carbon monoxide
for the reaction to occur. So this is what's kind of
capping off on how much this reaction can move forward. But the whole point of this was
to cancel that and that. So, obviously, if we're using
12.7 moles of carbon monoxide, we're going to produce
12.7 moles of methanol will be produced. And now we just have to figure
what is the mass of 12.7 moles of methanol. And we just think about what
the atomic weight. So if you look at methanol,
CH3-- let me put the H3-- H3OH. Its atomic weight is 12 plus 3
times 1 plus 3 plus 16 plus 1. So what is this? This is 20 plus 2 is equal to
32, or if we think in moler terms-- or not moler terms, I
should just say 32 moles-- this is its atomic weight. So that tells us that if we have
a mole of it we're going to have 32 grams of methanol
per 1 mole of methanol. And once again, we got
that by figuring out its atomic weight. Now to convert the number of
moles that we have of methanol to the number of grams, we just multiplied that times that. That units work out. This is in the numerator, this
is in the denominator. Let me just copy and paste it. So we have that, copy and
paste, and then we can multiply it times that. Let me copy and paste it. You get that right there, and
I'll pick a different color, maybe a blue. Just like that, and then let's
multiply these two. We have moles of methanol, and
we're left with 12.7 times 32 is equal to-- so let me just
keyboard again, let me clear it out-- 12.7 times 32, 406.4. I'm kind of adding an extra
significant digit there, but let's just go with it. 406.4-- actually, I
should just stay with significant digits. So we'll say 406 grams-- I'm
rounding down-- 406 grams of methanol-- let me write it with
the units-- so grams of methanol-- I went off the
screen-- of methanol produced. I could write the produced
down here since we're all done. So I think that was the first
part of our question. How many grams of methanol. What mass of methanol, that's
400-- I have a horrible memory-- 406 grams. And then
they say what mass of excess reactant remains after
the limiting reactant has been consumed? Now there's a couple of
ways you can do this. The easiest way to think
about it is that the mass has to be conserved. So we started off over here with
65 grams of-- let me be careful-- we started off here
with 356 grams of carbon monoxide, and 65 grams of
hydrogen, and we were able to produce 406 grams of methanol. And we figured out that
carbon monoxide was the limiting reactant. So all of this gets consumed,
and only some of this gets consumed. So if you do the math, what gets
consumed has to be equal to 406 grams because that's
what gets produced. So let's think about
it a little bit. The left-hand side, how many
total grams do we have? So if we have 356 plus 65, we're
starting off with 421 grams of reactant. So we're starting with 421 grams
of reactant, and then we end up with 406 grams of
product, of our methanol. So that means that 421
minus 406 grams of reactant was not used. So that means that 421 minus
406, which is equal to what? 21 minus 6 is 15 grams
of reactant not used. Now the reactant that's not
going to be used is the one that you have an excess of. And we figured out that we have
an excess of hydrogen. So all of that 15 grams must
have been 15 grams of hydrogen that was not used. So 15 grams of hydrogen
left over. Now the other way you could
have done this exact same problem is you could have said
look, we're starting with 12.7 moles of carbon monoxide--
that's the limiting reactant. It's a 2:1 ratio-- you need 2
moles of hydrogen for every mole of carbon monoxide. So you said, OK, if we have
12.7 moles of this, I need twice that many moles
of hydrogen. So you would say well,
what's twice that? That's 25.4. You need 25.4 moles, you have
32.5, so you subtract the difference-- you subtract 32.5
minus 25.4-- And that number of moles of hydrogen
is left over. You'd multiply it times the
grams per mole, which is 2, and then you, once again, would
get the same thing. You would get 15 grams. Let's do that. I just want to make sure
you understand. So we're starting off with 12.7
moles of carbon monoxide. And we know that we are required
that 2 moles of hydrogen are required for every
1 mole of CO, of carbon monoxide, that is required. So you know that this cancels
with that, you know 12.7 times 2 is 25.4 moles of
H2 required. And what is the mass of
25.4 moles of H2. So let's write that. 25.4 moles of hydrogen
required. Times-- well we know the
molecular weight of H2 is 2, so 1 mole, we're going
to have 2 grams of H2 per 1 mole of H2. This cancels out, mole
of H2, mole of H2. So we're going to have
25.4 times 2 is what? That is 50.8. That is going to be 50.8
grams of H2 required. That's how much we're going to
consume in the reaction. Now they wanted to know
how much is left over. So we're going to
consume 50.8. We started with 65. So if you subtract 50.8 from
65-- 65 minus 50.8-- you're going to get what? You're going to get 14.2
grams left over. And that compares to the 16 we
got left over when we just took the 421 minus the 406. And the difference between the
14.2 and the 15 is really just a little rounding here and
there with our digits. Anyway, hopefully you
found that helpful.