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# Stoichiometry: Limiting reagent

## Video transcript

Let's say we have some
ammonia gas. That's NH3 and it's a gas,
that's why the g is in parentheses. And we combine that with some
oxygen, molecular oxygen, it's also a gas. And that reaction produces
some nitrogen monoxide. NO, monoxide, there's only
one oxygen there. That's also a gas, it's also
called nitric oxide, not to be confused with nitrous oxide. I'll write nitrous oxide, it's
N2O, this is laughing gas. Anyway, I don't want to divert
you too much, we have to focus on the problem. So this is nitric oxide
or nitrogen monoxide. It's a pollutant, I think
it comes out of some cigarette smokes. I think it's used in
the body, as well. You could we could look
it up on the internet. There's your oxide. Ammonia is an important
fertilizer. You know all about oxygen. And this also yields
some water. So, plus some H2O. And we're told that we're given
34 grams of ammonia. And we're given 32
grams of oxygen. This is going to be the
oxygen molecule, O2. So the question is, how many
grams of nitrogen monoxide, or the nitric oxide, are going
to be produced? So how much of just the NO is
going to be produced in grams? So this is a stoichiometry
problem. And so the important thing first
is to just make sure we have a balanced equation before
we even start anything. And lo and behold, we don't
have a balanced equation. Let's confirm that it's
not balanced. Let's see, we have one nitrogen
here, we have one nitrogen there. That looks balanced so far. And remember, the pattern is
start with the complicated stuff, leave the single atom
molecules for last. Because those you can fix at the
end without missing anything else up. Hydrogen, we have three
hydrogens on the left-hand side. On the right-hand side we
have two hydrogens. So let's see, how can we have
three hydrogens on the right-hand side? If we multiply this times one
and a half, 1.5, now we have three hydrogens on this side. 1.5 times 2, we have
three hydrogens on the right-hand side. Things are looking good. We have two oxygen on
the left-hand side. How many oxygens do we have
on the right-hand side? We have one oxygen here, and
we have one oxygen in this molecule, but we have one and a
half of the whole molecule. So we have one and a half
oxygens and then we have one more oxygen. So we have two and a half
oxygens on the right-hand side, we only have two
in this molecule. So what do we have to do? How can we get two and
a half oxygens here? Well if we multiply it
by 5/4, or 1.25. 5/4 times 2 is 5/2,
which is 2.5. So now we have 2.5 oxygens. 5/4 times 2 is 2.5. 1 plus 1.5 is also 2.5. Well, we're not balanced yet. We can't leave this equation
with just these weird decimal numbers over there. So let's rewrite it. If we wanted to get rid of all
of these, we can multiply the entire equation by 2. No, not 2. We have to multiply the entire
equation by 4 to get rid of this 4 in the denominator. So we multiply the entire
equation by 4, we have 4 molecules of ammonia. We can even think of it
in terms of moles. Right now I'm thinking of
individual molecules. But we could say we have
4 moles of ammonia. 4 times 6 times 10 to the
23 molecules of ammonia. Either way it all works out. Hopefully, you're starting to
see the value of moles. Plus 5 molecules, I'll just
think of it in terms of individual molecules for
now, plus 5 molecules of molecular oxygen. We're just multiplying
everything by 4, that's what we're doing. Yields 4 moles of nitrogen
monoxide. So 4NO plus, we multiplied
both sides by 4, so plus 6 waters. 6 H2O. So there you go, we
got some good practice balancing equations. So let's go back to the
original problem. We're given 34 grams
of ammonia. So what we need to figure out
is how many moles of ammonia were we given? So what's the atomic
mass of ammonia? Not the atomic mass, what's the
molecular mass of ammonia? We're dealing with nitrogen
14, so it has a mass number of 14. Hydrogen has a mass
number of 1. So each nitrogen has
a mass of 14. And then the hydrogens each
have a mass of 1. Remember, hydrogen is
kind of strange. It doesn't have neutrons,
or at least in its most traditional form. So it has just a mass
number of 1. It's just a proton and an
electron if it's neutral. So this is 3 times 1. We have three hydrogen atoms. So
the mass, the atomic mass, of one molecule of ammonia
is 14 plus 3 is 17 atomic mass units. Or, another way to write that,
if one molecule of ammonia is 17 atomic mass units,
then 1 mole of ammonia is how many grams? It's going to be 17 grams. So how many moles of ammonia
are we given? We're given 34 grams of ammonia,
1 mole is 17 grams. So we're given 2 moles. 34 is 2 times 17. So this is 2 moles. We're given 2 moles
of ammonia. Let's see how much oxygen
we're given. Or how much of the molecular
oxygen we've been given in this case. So let's see, the mass of just
oxygen by itself is 16. The mass of just the atomic
oxygen, you have to be a little bit careful here because
sometimes people say we have 32 grams of oxygen when
they're really talking about the molecular oxygen. Well, I guess it doesn't
matter either way. But, sometimes when they talk
about oxygen you have to make sure whether it is molecular
or atomic oxygen. But the atomic mass number
of oxygen is 16. I can confirm that
by looking at the periodic table down here. So what's the molecular mass of
the diatomic molecule O2? Well it has 2 oxygen, so it's
going to be 2 times 16 equals 32 atomic mass units. One molecule of O2 is 32
atomic mass units. Or 1 mole of O2 is
how many grams? Well if one molecule is 32
atomic mass units, then 6 times 10 to the 23 of that
molecule are going to be that many grams. 32 grams. So how
many moles of oxygen have we been given? We've been given exactly 32
grams of oxygen, which is exactly 1 mole. So we've been given 34 grams of
ammonia, which is 2 moles. Let me write that in a
nice vibrant color. And we've been given 1 mole
of the oxygen molecule. Now, when we look at this
reaction, for every 4 moles of ammonia, we need 5
moles of oxygen. Or for every 5 moles of oxygen
we need 4 moles of ammonia. So something doesn't gel here. Normally, we need more
moles of oxygen than we have ammonia. In the example that we're
working through, we've been given less moles of oxygen
than ammonia. We've been given less
oxygen than we need for all of this ammonia. In an ideal world, if we had 2
moles of ammonia, we would need 2.5 moles of oxygen. The ratio of ammonia
to oxygen, let me write that down. In a different color, this
is getting boring. The ratio of ammonia, NH3 to
oxygen in our balanced equation, let me put that in a
square so you know this is the most important part of
what I'm writing. The ratio in this
reaction is 4:5. So if I'm given 2 moles
of ammonia, this is equal to what? If I'm given 2 moles of ammonia,
how many moles of oxygen do I need? I need 2.5 moles of oxygen. Whatever this is,
is that right? Yeah, 4/5. 1.25. If you doubled both of these
numbers, you get 4/5. So I need 2.5 moles. But I don't have 2.5
moles of oxygen. I only have 1 mole of oxygen. So oxygen is going to
be the limiting reagent in this reaction. I don't have enough oxygen. I have plenty of ammonia, but I
don't have enough oxygen to react with it. So this is the limiting
reagent. And I said before, the word
reagent and reactant are used interchangeably. But when people talk about the
limiting ones, they tend to call it the reagent. So oxygen is the limiting
reagent. So we have extra ammonia. So given that we have 1 mole of
oxygen, how many moles of ammonia can I react with that? So this reaction is going to
look something like this. I only have 1 mole of oxygen. So instead of 5O2, I
have to write 1O2. Let me make sure that's
not of 10. Let me do it in a
different color. I only have 1O2 instead
of 5O2. So how many ammonia are going
to react with that? Well the ratio is 4:5. So I'm going to have
0.8 ammonia. 4 is to 5 as 0.8 is to 1. And so essentially, if I take
this whole equation up here and divide it by 5,
I'll get what's actually going to happen. So this divided by
5 is 0.8, nitric oxide or nitrogen monoxide. Plus 6/5 moles of H2O. And so the original question in
the beginning is, how many grams of nitric oxide
are we going to produce, or nitrogen monoxide? So we have 1 mole of oxygen,
0.8 moles of ammonia, and we're going to produce 0.8 moles
of nitrogen monoxide. Because we only have
one oxygen. So 0.8 moles of nitric oxide
or nitrogen monoxide. So what's the atomic
mass, the molecular mass, of nitrogen monoxide? Nitrogen has 14 atomic
mass units. Oxygen is 16. We've done that before. But you can confirm, your oxygen
is 16, nitrogen is 14. So one molecule of nitrogen
monoxide is equal to 30 atomic mass units. 1 mole or 6.02 times 10 to the
23 molecules of nitrogen monoxide, therefore, will be 30
grams. And how many moles are we producing in
this reaction? Because oxygen was the limiting
reagent, we only had 1 mole of oxygen here. Because of that, we can
only produce 0.8 moles of nitrogen monoxide. So 0.8 moles of nitrogen
monoxide, 1 mole is 30 grams. So 0.8 moles of NO is going to
be equal to 0.8 times 30, which is equal to 24 grams. So
we're going to be able to produce 24 grams of
nitrogen monoxide. And so you might ask a question,
we're only using 0.8 moles of ammonia. We were given, in the original
problem, 2 moles of ammonia, so what happens with all
the leftover ammonia? Assuming we mix it really good,
we're going to literally end up with 1.2 moles of
ammonia just doing nothing at the end. So we're going to have 24
grams of nitric oxide. And then we used 0.8
moles of ammonia. And we're going to have left
1.2 moles of ammonia. And if you want, you can figure
out how many of the original grams of
ammonia that is. You just figure out how many
grams 1 mole of ammonia is, it's 17 grams. And then just
multiply that times 1.2. Anyway, I hope you found
that interesting. If you found this a little bit
confusing, just watch the video again and pause it and
try to solve it yourself. But I think it should start
making a little bit of intuitive sense. I think the hard part is just
the conversions between moles and grams and getting
that part right. And then making sure that you
understand the ratios. You understand this is a 4:5
ratio, there's always going to be less ammonia than oxygen. And if I only have 1 mole of
oxygen, I'm going to have to have less than 1 mole
of ammonia, even though I was given 2. So oxygen, in this example, was
the rate limiting factor. Let's say I had 10 moles of
oxygen and 2 moles of ammonia. In that case, ammonia
would have been the rate limiting factor. Because then I would have had
more than enough oxygen. Because for 2 moles of
ammonia, I only need 2.5 moles of oxygen. So 10 moles would have
been overkill. Anyway, see you in
the next video.