Mr. Andersen explains the concept of a limiting reactant (or a limiting reagent) in a chemical reaction. He also shows you how to calculate the limiting reactant and the percent yield in a chemical reaction.
Transcript Provided by YouTube:
Hi. It’s Mr. Andersen. And today I’m going to be talking about limiting reactants and
percent yield. Sometimes we call limiting reactant limiting reagents. But those are
essentially the same thing. And this is always going to occur in a chemical reaction. So
before we talk about chemical reactions, let’s actually talk about food. And so if we were
making cookies, let’s say we were making Nestle’s Chocolate Chip Cookies. This is the recipe
for it. And if you were to go down your ingredients list before you start making the cookies and
you see, yeah I’ve got the flour. I’ve got the teaspoon of baking soda. I’ve got 3/4
cup of sugar. But then you get down here and you look in the fridge and you realize, oh
I only have one egg. And I need to two eggs for this recipe. Well chemical reactions work
the same way. You’ve got reactants. And you can think of those as like the ingredients
before the actual reaction. And if you don’t have all of them, then there’s going to be
one thing that’s going to limit that reaction. We call that the limiting reactant. In this
case it would be eggs. Now can I not make cookies anymore? No way. All you do is just
cut all of these down. So if they only have one large egg instead of two, then I just
cut everything in half. So it’s a half of teaspoon of salt. A half a cup of butter.
And so it still works. I’m still able to make cookies. But I just cut everything in half.
And so those coefficients that you see in front of equations, if you think of those
as like the amount is a good way to think about it. So let me kind of start by showing
this in a visual sense. So if we’re doing hydrogen combustion and an example of this
would be, this is in the main rocket chamber in the space shuttle. What we have in here
is liquid hydrogen and liquid oxygen. And those are combining and we’re getting this
huge exothermic reaction which is creating a lot of the lift. So this would be the chemical
reaction. Hydrogen plus oxygen yields H2O. If we actually draw that out then, here would
be the reactants on the left side. This would be hydrogen gas and this would be oxygen gas.
And then the products I haven’t shown you yet. And so if you look at this, you know
that H2 and O are going to combine together to make H2O. And so you should, if you pause
the video right now, you’d be able to take a second and probably figure out what the
limiting reactant is. Looks like we’ve got a little more hydrogen then oxygen. So what’s
the limiting reactant? Let’s do the actual reaction. So what do we get? We get 4 water
molecules. And then we have 1 oxygen left over. And so what was the limiting reactant?
The limiting reactant in this case was the hydrogen gas. Because how many atoms of hydrogen
gas did we have? We’ve got 2, 4, 6, 8. And so we only had enough to make four water molecules.
Now we had enough oxygen to make 2 more water molecules. But the hydrogen in this case were
like the eggs in our recipe. They were limiting the reaction. Let’s go to the next one. This
is combustion of methane. Combustion of methane, what you’re doing in that is you’re actually
combining natural gas. And that’s, methane is a hydrocarbon. So it’s 1 carbon and 4 hydrogens.
With oxygen just found in the room and we’re making carbon dioxide and water. That’s how
a bunsen burner works. So if we show you the reactants that I happen to have in this really
small area. I’ve got 1 methane molecule. And then we’ve got, it looks like 4 oxygen molecules.
And those are going to make water and carbon dioxide. So if you actually do the reaction,
there we go. I get 2 water molecules and I’ve got 1 carbon dioxide molecule. So what was
the limiting reagent this time? The same thing. Again we’ve got plenty of oxygen right here,
but we’ve only got 1 methane molecule. And so what was limiting us was the methane. So
we call that the limiting reactant or the limiting reagent. So what would be left over?
Well everything that didn’t’ react. In other words that oxygen that was left over was just
going to stay as oxygen. It’s not used. It’s like the other half a cup of butter that we
used in our recipe. And so those are graphically kind of showing you what limiting reagents
are. Unfortunately atoms aren’t this big in the laboratory. If they were this big we could
just look at how much we have and we’d be able to figure out which is going to be the
limiting reactants. And so we have to use our friend the mole to actually figure out
how much there is. And so let’s say we have this reaction down here. We’ve got methane
plus oxygen yields carbon dioxide plus water. And let’s say together I’m going to mix two
things. I’m going to take 10 grams of oxygen and I’m going to mix that with 10 grams of
methane. So the question that you could be posed with is which of those is the limiting
reagent? Which of these is the limiting reactant? Is it the methane or is it the oxygen? Well
you can’t tell by looking at a mass. So you have to do some calculations. And we can’t
see the atoms so we’re kind of stuck. And so let me talk you through this process. First
thing we do is we convert them to moles. We then compare the moles. In other words we’re
going to compare the amount of methane and oxygen to one of the products on the other
side. In this case I’ve chosen carbon dioxide. We’re next going to convert that back to grams.
And then we’re going to figure out all the possible products. And so if these steps are
confusing to you, you may want to look at the video on converting grams to moles. So
first thing we do is convert to moles. And so I’m going to put my grams on the bottom
and then my moles on the top. These numbers right here are the formula masses. And so
we’re getting that from the periodic table. In other words to get methane we’re taking
the atomic mass of 1 carbon plus 4 hydrogens to get that. Okay. So if we look at it what
we’ve converted to, if I were to cancel out my grams of methane, is I’ve converted to
moles of methane and moles of oxygen. If we do our next conversion I can then compare
the moles. And so on the bottom here I have 1 mole of methane. I’m grabbing that from
the top over here. And one mole of oxygen to two moles of oxygen. Now where am I getting
2 on the bottom and 1 mole on the top? I’m doing that from my actual equation here. And
so there’s an implied 1 in front of the methane right here. And so that’s why there’s 1 mole
here. And there’s 2 moles of oxygen. Why did I only put 1 mole of carbon dioxide on the
top? Because on the other side, on my product side, there’s only 1 mole. Now we could have
chosen instead on this comparing moles, we could have put one mole of methane on the
bottom and 2 moles of water. In other words we could have compared both of these reactants
with water. And we still would have figured out which is the limiting reactant. Next I
could convert it to grams. And I do that just by now taking the moles of carbon dioxide
and figuring out the formula mass of carbon dioxide. And then I finally get my possible
results. And so if I had 10 grams of methane. That’s enough methane to make 27.4 grams of
carbon dioxide. So that’s a lot. That 10 grams of oxygen however could only make 6.88 grams
of carbon dioxide. And so what’s the limiting reactant? Well this equation or these equations
or these conversions show me that I’m going to run out of oxygen long before I’m gong
to run out of methane. And so what’s the limiting reactant? The limiting reactant is the amount
of oxygen. In other words 10 grams of oxygen only gets me 6.88 grams of carbon dioxide.
And so that’s the limiting reagent, the amount of carbon dioxide. And I can even predict
how much grams of carbon dioxide I’m going to get. So if you think of yourself as a chemist,
what you could now do is put in a 6.88 grams of carbon dioxide. You could work backwards.
And you could figure out exactly how much methane you have to add to get 6.88 grams
of carbon dioxide. And if you’re a smart chemist you’re always in any reaction that you do,
you’re always going to make your most expensive chemical as a reactant be your limiting reactant.
So you’re not going to send, you’re not going to use any bit more of that. Okay. The other
thing I mentioned I would talk about is the percent yield. And so in this theoretical
experiment we’re going to get 6.88 grams of carbon dioxide. That’s with all the oxygen
reacting with all of the methane. But that never happens. So this is a what we call,
6.88 would be a predicted yield. If everything went perfect. In other words if every molecule
of methane reacted with every molecule of oxygen we’d get 6.88 grams of carbon dioxide.
Now you can think about it. That’s never going to happen if you think about the billions
and billions and billions of molecules. The chances that they’re all going to react and
some aren’t going to be lost is really, really rare. And so lots of times you’re actual yield
will be much less than your predicted yield. In other words I might be hoping to get the
6.88 but I only get 6.19. And so those 2 values can tell me the percent yield. And so the
percent yield is essentially the actual yield. In other words how much I get divided by the
predicted yield, how much I was hoping to get. And so if I throw those values in here,
you can see the smaller value is going to be on the top. That’s how much I actually
get back. And the precent yield is never 100 percent. And if I divide the top by the bottom
and then multiply it by 100 percent I get 90.0 percent, percent yield. And that show
me that this is a pretty good reaction. In other words I’m getting most of the product
that I predicted that I would be able to get. And so that’s limiting reactants. That’s percent
yield. And I hope that’s helpful.
This post was previously published on YouTube.
Photo credit: Screenshot from video