Learning chemistry can sometimes feel like running an obstacle course — there’s a bunch
of topics to go through, and they all require different things.
Luckily, at this point in the series, we’ve learned a lot of those skills.
So far, we’ve talked about reaction types, balancing equations, the mole, stoichiometry,
and limiting and excess reactants.
When talking about calculating how much of the excess reactant is left over, we’ll
need to rely on all of these topics to get to the finish line.
It sounds like a lot, but don’t let that intimidate you.
We’ve come a long way since episode one, building up confidence to leap over the hurdles
that used to make us stumble.
Besides, calculating how much of a substance is left in excess is a useful skill we’ll
need to know in all kinds of industries beyond just chemistry.
It even drives the business decisions people make every day to cultivate a healthy economy.
Hi, I’m Will Comar, and welcome back to Study Hall Chemistry, presented by Arizona
State University and Crash Course.
Let’s dive in!
Understanding how much substance is left over is as much an economical concept as it is
a chemical one.
Grocery stores that purchase bulk produce from a supplier need to know how much to buy.
If they get too much and can’t sell it all, the food will eventually spoil, and the whole
thing ends up being a waste of money.
Having just the right amount for the number of customers ends up being more cost efficient
and more sustainable.
In chemistry, it’s uncommon for reactions to have exactly the right amounts of reactants
to make a product.
Sometimes, we have too little or too much.
There’s usually one that gets used up first, which we call the limiting reactant,
and one that has some left over after the reaction is complete, called the excess reactant.
When reactions don’t use up all of a substance, especially if that substance is rare or expensive,
it’s a waste of that reactant — and sometimes the amount of product it makes,
called the yield, isn’t worth that waste.
Luckily, there’s a good way to avoid this. We can use stoichiometry to calculate how
much of an excess reactant will get used up,
and how much will be left over.
We can figure out if the reaction will give us an efficient yield before we invest in the reactants.
It’s not as simple as just subtracting the grams of our products from the grams of our reactants, though.
To think about why this is, let’s use some algebra.
If we try to calculate 3y-2x, we can’t just subtract 3 minus 2.
We have to work around our variables, x and y.
But if we know that 2x equals y, we have a way of translating the problem so both our
values are in terms of y.
Now that they’re speaking the same language, we can subtract and get 2y as the answer.
Finding excess amounts in chemistry works in the same way.
We can’t just subtract grams of different substances from each other to get what’s left over.
They have to be translated, or expressed in terms of the same units and the same substance.
Let’s try this with an example.
Aluminum chloride is commonly used in small amounts for antiperspirants, and has the chemical formula AlCl3.
If you remember from when we talked about this back in episode 4, it’s produced by
combining aluminum and diatomic chlorine.
If we wanted to produce some aluminum chloride to sell to a deodorant manufacturer,
let's find out how much excess reactant we’d have.
To do that, we need to calculate how much aluminum chloride we could produce with 122
grams of aluminum and 239 grams of chlorine.
This will tell us what our limiting and excess reactants are. From there, we can calculate
how much excess there will be.
We can follow the same steps we covered last time on how to calculate the theoretical yield of a reaction.
First, we’ll use our given amounts of aluminum and chlorine to find how much product each
could make on their own.
We’d divide 122 grams of aluminum by its molar mass.
This converts us from grams to moles.
From here, we need to multiply by the mole ratio we get from our reaction coefficients
to convert from moles of aluminum to moles of aluminum chloride.
We know that it takes 2 moles of aluminum to make 2 moles of aluminum chloride, so our ratio is 2:2.
Then all we need to do is multiply by the molar mass of aluminum chloride to get from
moles to grams, which gives us the amount produced by aluminum individually at 596 grams.
Then, we’d do the same thing for 239 grams of chlorine, using a different mole ratio,
and then multiply by the molar mass of aluminum chloride to convert to grams.
It turns out that 300 grams of aluminum chloride is produced by chlorine alone.
When we compare our two results, 300 is less than 596, which makes 300 grams of aluminum
chloride our theoretical yield, and chlorine our limiting reactant.
Aluminum is our excess reactant, so now we need to find out how much excess aluminum we have left over.
Like I mentioned, we can’t just subtract grams of aluminum from the reaction’s aluminum chloride yield.
To find our excess aluminum, we need to work in terms of grams of aluminum only.
So, what we need to do is add one more stoichiometry step to our calculation, complete with its own mole ratio.
We need to convert the given amount of our limiting reactant to grams of our excess reactant
to find out how much excess reactant was consumed in the reaction.
Since chlorine is used up first as the limiting reactant, it dictates how much aluminum gets spent.
We need to put grams of chlorine in terms of aluminum,
so we can find out how much aluminum was consumed in the reaction.
From there, we can subtract how much was consumed from our given amount of excess reactant to
get how much is left over.
This process will require a couple conversion factors.
We’ll start by converting grams of chlorine to moles, with the molar mass of chlorine.
Next, we need a mole ratio that converts from moles of chlorine to moles of aluminum.
To cancel moles of chlorine, we should put them on the bottom and moles of aluminum on top.
From here, all we need to do is convert from moles to grams, using aluminum’s molar mass.
We end up with 61.5 grams of aluminum, which is the amount consumed in the reaction before
we run out of chlorine.
So, now that we know how much aluminum is used up, we just have to subtract it from
how much we started with to get how much is left in excess.
Now we can report to the deodorant company that there are 60.5 grams of aluminum left
over when our reaction is complete.
No sweat!
Let’s try another example with an acid-base reaction.
These kinds of reactions produce water and a type of salt.
Let’s say we work in a chemistry lab where someone accidentally spills 0.115 liters of hydrochloric acid.
One way to resolve the problem would be to apply calcium hydroxide, which is a basic substance, to it.
Since acid-base reactions produce water, a neutral substance, adding a base to an acid
can neutralize it so it isn’t dangerous to the people in the lab.
We’ll cover neutralization in more detail next time, but for now, we can focus on the
excess reactants perspective.
If we have excess acid left over, it means our neutralization won’t be complete.
So, we’re asked to find out whether or not 0.049 liters of calcium hydroxide is enough
to fully resolve the spill.
In other words, how much excess reactant we’d have left over.
To figure this out, we’ll need to apply a bunch of our skills in one big mega-problem —
let's treat it like an obstacle course.
We already know that this is an acid-based reaction, so we can make the leap
and predict that the products need to be water and a salt — calcium chloride.
Now we can add our states of matter and balance the equation, adding coefficients where needed.
Our next obstacle is to find out what our limiting and excess reactants are by calculating
the theoretical yield.
Our reactants are liquids, but let’s swap liters for grams to make our calculations easier.
One liter is equal to 1000 grams, so we’ll multiply our givens by 1000.
Now we’ve got 115 grams of hydrochloric acid, which we can convert to moles using molar mass.
We’ll use a mole ratio to convert to moles of calcium chloride, and then the molar mass
to convert to grams.
It looks like hydrochloric acid alone produces 176 grams of calcium chloride.
If we do the same calculation setup for 49.0 grams of calcium hydroxide, we get 72.8 grams
of calcium chloride.
Since calcium hydroxide produced far less calcium chloride, it is the limiting reactant,
and hydrochloric acid is the excess reactant.
Our expected, or theoretical yield, is the smaller of the two numbers, at 72.8 grams.
Now let’s finish our race with the calculations of excess hydrochloric acid consumed and left
over with some help from TestTube.
For this leg of our race, we’ll begin with our given amount of our limiting reactant,
since this is what’s controlling how much of our excess reactant actually gets used.
So we start with 49.0 grams of calcium hydroxide, and we’re off!
Let’s first use its molar mass to convert from grams to moles.
Once we’ve divided that and cancelled our grams, we can move on to our mole ratio.
We need to convert from moles of our limiting reactant to moles of our excess reactant,
hydrochloric acid.
From our coefficients, we determine the ratio is two moles of hydrochloric acid per mole
of calcium hydroxide.
Now that we’re in terms of the same substance, we can use the molar mass of hydrochloric
acid to convert from moles of our excess reactant to grams.
This will tell us how much hydrochloric acid was consumed in the reaction.
Since the spill of hydrochloric acid was 115 grams, we can subtract our amount consumed,
47.7 grams, to find out that our amount of excess reactant left over is 67.3 grams.
We’re almost to the finish line — we just need to put our answers in context.
Since we have leftover acid, we won’t be able to completely neutralize the spill. Yikes!
If you’re wondering how chemists know exactly what we need to neutralize the spill,
we’ll talk about this when we cover neutralization and titration.
For now, though, we can at least tell we’ll need to use more calcium hydroxide in the
reaction we’ll actually perform.
And we’re done — that gold medal is ours!
This example illustrates why it’s important to calculate how much our amounts will produce
and use up before mixing our chemicals.
If we just dove in before doing the math, we wouldn’t have fully cleaned up the spill
and our effort would have been kind of a waste of time.
The importance of this goes far beyond lab safety, too.
There are all kinds of industries that rely on proper amounts to execute their reactions
to produce what they need and to avoid wasting materials.
From environmental remediation of mining pollution to developing cost-effective steel for a construction
project, excess reactants pop up everywhere.
Next time, we’ll continue learning about neutralization with an introduction to pH
and more complex acid-base relationships.
We’ve only got two more episodes left — hopefully at this point in our journey you’re starting
to look forward to our next topics as much as I have!
See you all then!
Thanks for watching Study Hall: Chemistry, which is produced by Arizona State University
and the Crash Course team at Complexly.
If you liked this video and want to keep learning with us here in Study Hall, be sure to subscribe.
You can learn more about ASU and the videos produced by Crash Course in the links in the description.
See you all next time!