>> Welcome to a screencast on an introduction to limiting reactants.
Okay, for our next example, we're also going to be using a reaction that involves iron
and oxygen, but this time it's the reaction of iron with oxygen to make Fe2O3.
And we're going to look at this pictorially, sort of on a molecular or atomic level.
And first of all, we need the chemical equation.
And when it's written out in balance, four Fe react with three O2 to make two Fe2O3.
And with the amounts we are given, here is what that would look like.
We have eight moles of Fe, so each of these blue spheres represents,
on an atomic level, one atom of Fe.
On a molecular level -- or on a larger scale, a level that would represent a mole of Fe.
And then we have seven moles of O2.
Remember that O2 is diatomic -- two oxygen atoms bonded together.
So we have seven groupings of pairs of oxygen atoms.
So that's seven moles of O2, or on a smaller scale, seven molecules of O2.
And then we're going to see what happens when these substances combine to make Fe2O3.
So see if you can predict what would happen.
Move the atoms around so that they end up joining as two --
sorry, as two irons and three oxygens stuck together in a unit.
And if you do this, let's kind of look at it here, with a couple things to note.
One is eight moles of Fe.
That's kind of straight forward.
But seven moles of O2 is, of course, the same as 14 moles of O. You have seven pairs
of oxygen atoms, therefore you have 14 actual atoms.
And on a larger scale, 14 moles of O. And so we're going to have to end
up with 14 moles of O and eight moles of Fe.
And if we now do this, well, there's an Fe2O3 unit.
There's another Fe2O3 unit.
And another.
And keep track of how many Fe we are using and how many O2.
And you notice, when we get to this point, we've used eight Fe -- so these are all gone.
And we've used one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, of the oxygens.
And we've run out of iron.
There's no more iron left to react.
So the oxygen -- there's a little bit of it left over -- didn't react with anything.
So what's left in our container is going to look like this.
This is our product's reaction if it's 100% reacting.
So note that when this reaction takes place, what happens is all of the iron gets used up.
And there's some leftover O2.
And this is the kind of thing that can happen sometimes.
What we would say is the iron is the limiting reactant.
We ran out of it.
It limits how much product can be formed.
And we would say that oxygen, O2, is the excess reactant.
There's more of it than we need and it doesn't limit how much product can be formed.
So whenever you're dealing with quantitative aspects of chemical reactions,
it's important to pay attention to whether or not there's a limiting reactant
because that's going to affect how much product can be formed.
So let's go back to the problem we've been dealing with and let's do the next part.
First of all, how many moles of Fe2O3 could be produced?
And it's very important here to note that whenever you do calculations
when you have a limiting reactant, you need to start your calculations
with the limiting reactant because that's what determines how much reaction can take place.
So in this case, we have eight moles of Fe.
We know from the chemical reaction that for every four moles of Fe that reacts,
two moles of Fe2O3 can be produced, so that's our stoichiometric conversion factor
that gives us that four moles, 4.00 moles, of Fe2O3 can be produced.
That's our theoretical yield, our maximum yield from this amount of Fe.
Now next to determine how many moles of O2 are left over, what we need to do in this case is,
once again, we start with our limiting reactant, eight moles of Fe.
And now we note that four moles of Fe require three moles of O2 to fully react.
And what that means, then, is that six moles of O2 are needed to completely react
with the given known amount, four -- sorry, eight moles of Fe.
And now if we look at the moles of O2 needed and compare that to the moles of O2 available
and note we have more moles of O2 available than --
than we need, then the difference between those two is going to be our excess.
So 7 minus 6 moles is 1 mole of O2.
That's our leftover.
So our last example for this screencast is a reaction between zinc and hydrochloric acid.
And we're asked -- how much hydrochloric acid is needed
to completely react with a given amount of zinc?
Notice, there's something a little different here.
We're determining the volume of a three molar hydrochloric acid solution,
so this is going to be similar to previous calculations,
but there will be the difference of, instead of using a grams and moles calculation
or conversation, we'll be using a volume and molarity conversion.
So we start with what we have -- 25.548 grams of zinc.
We need to convert that to moles, and one mole of zinc is 65.39 grams.
And then from the stoichiometry of the reaction, we know that for every one mole
of zinc, it requires two moles of HCl.
And then we're going to have to use our molarity to convert from moles of HCl to liters
of HCl solution and we know that there's three moles of HCl in one liter of solution.
And so we string the calculations one after another.
Grams of zinc cancels.
Moles of zinc cancels.
Moles of HCl cancels.
Leaves us with liters of HCl solution.
Or if you prefer it in milliliters, 260 milliliters of HCl solution.
And that's it for an introduction to limiting reactants.