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# Another stoichiometry example in a solution

Video transcript

Let's do another stoichiometry
problem where the reaction is occurring inside
of a solution. So we have a metallic zinc and
it reacts with aqueous hydrochloric acid. We see that right there, the
metallic zinc. And then the hydrochloric acid is
dissolved in water, it's an aqueous solution. The solvent is water. Makes these products
right there. And they say, what volume of
0.25 molar-- so they're giving us the concentration
of the solution. What volume of 0.25 molar
hydrochloric acid in milliliters-- so we've got to
remember that, they want their answer in milliliters-- is
required to convert 11.8 grams of zinc completely
to products? So the way we can do this
we'll figure out. We know that we're
starting off with 11.8 grams right there. And we just have to figure out
well, if we have 11.8 grams of zinc, how many moles
of zinc that is. And then, assuming this is a
balanced equation-- and I'm going to verify that right after
I finish this sentence-- but assuming that this is a
balanced equation, for every mole of zinc we're going
to need two moles of hydrochloric acid. And then we can figure out
given how many moles of hydrochloric acid we need, then
how much of the solution do we need based on
its concentration? So let's do that. It's going to, I think,
be a lot simpler when we actually do it. But I just want to give you
the action plan before we break into it. But before we do anything,
let's verify that this is really a balanced equation. So you have 1 zinc on the
left-hand side, 1 zinc on the right-hand side, 2 hydrogens
on the left, 2 hydrogens on the right, 2 chlorines
on the left, 2 chlorines on the right. So it looks balanced. Let's do the problem. So the first thing we want to
do-- we're starting off with 11.8 grams of zinc, they give
us that information. Now we're going to figure
out how many moles of zinc that is. We're going to want to convert
this to moles. So it's going to be moles
of zinc over grams of zinc moles per gram. And if you look it up on a
periodic table, zinc has an atomic weight of 65.38, so
its molar mass is 65.38. So it's 1 mole for every 65.38
grams, or its 1/65, roughly, of a mole per gram. Actually, you know what? I'm just going to save our
calculator for the end. So the grams of zinc cancels
that with the grams of zinc, and we're left with 11.8 divided
by 65.38 moles of zinc. Now we're trying to figure out
how much of this solution of hydrochloric acid is required. So the next thing we want to
figure out is how many moles of hydrochloric acid
are required. And we see that for every mole
of zinc we need 2 moles of the hydrochloric acid. Let me write it this way. For every 1 mole of zinc--
we'll put that in the denominator. I'll switch colors just
arbitrarily here-- we need 2 moles of hydrochloric acid. That is required. And so then we can multiply
the moles of zinc. We'll cancel out the moles of zinc.
That's why I wrote 2/1 instead of 1/2. Two moles of hydrochloric acid
for every mole of zinc. And what does this give us? This gives us-- I'm not even
going to do the math. It's 11.8 times 2 over 65.38. And then we're left with moles
of hydrochloric acid required-- moles
of hydrochloric acid-- HCl-- required. So we haven't solved
the problem yet. We want to figure out
what volume of the solution we need. We know the actual number of
molecules we need, but we need to know how much solution
will contain that number of molecules. So we want to express
this in liters. Well, eventually we're going
to get to milliliters, but let's just go straight to liters
first because molarity is always expressed in liters. So let's multiply this. We want the moles of
HCl to cancel out. So let's put that in the
denominator, moles of hydrochloric acid. And we want liters of
hydrochloric acid, let me put it this way, liters
of 2.50 molar hydrochloric acid solution. When anyone puts this big M
there, that's molarity. You assume they're talking
about a solution because they're giving you the
concentration of the solute in the total solution. So how many moles are
there per liter? Well this concentration
is telling us. Molarity is moles per liter. So there are 2.50 moles for
every 1 liter of our solution. That's what that tells us. This is a 2.5 molar solution. Its molarity is 2.5. There are 2.52 moles
for every liter. And I wrote the moles in the
denominator because I wanted to cancel out here. So it's 1/2.5. So let's cancel those
two guys out. That cancels out with
that right there. And then we are left with 11.8
times 2 in the numerator. And in the denominator, we have
65.38 times 2-- let me write the times not as
a-- I'll write it that way-- times 2.50. And all we're left with in our
units is liters of 2.50 molar HCl solution-- and then in
that green-- required. We're almost done. Then we can get break out
the calculator and get the exact answer. They want the answer
in milliliters. So far we have the
answer in liters. I mean, there's going to be some
number here, I don't know what it is. But it's that number of liters
of the 2.5 molar hydrochloric acid solution is required. Let me just copy
and paste this. Let's convert this to
milliliters because that's what they want. So that's what we had from
the previous line. And to convert this to
milliliters-- we've done this drill before-- we want the
liters to cancel out. I'll just do it like this. And we want our answer
in milliliters. And how many milliliters
are there for a liter? Well there are 1,000 milliliters
per 1 liter. And all of these steps-- the way
I'm showing it to you, I'm making the dimensions
cancel out. But hopefully, they're making
some intuitive sense as well, that we divide by 2.5 at this
stage because we have more than 1 mole per liter. So however many moles we
require, we're going to have fewer liters required. So we're going to be dividing
by a number that's greater than 1. If that doesn't make sense to
you, just stick to the units. But at every one of these steps
I really want you to actually conceptualize
the numbers as well. But let's go back
to the problem. So we're converting
it to milliliters. So our final solution we're left
with, we have the liters. Cancelling out the liters, and
so we're left with, in the numerator, 11.8 times 2 times
1,000 over 65.38 times 2.50. And in the numerator we're left
with milliliters of 2.5 molar hydrochloric acid. I'll just write it milliliters
of 2.50 molar hydrochloric acid. I don't want this 2.50 molar
HCl to confuse you. These aren't numbers in the
computation, this is a label to say what type of solution. We could have just written
over here liters of the solution in question. In case these numbers confuse
you, we could have written liters of the solution. And then this would be
milliliters of the solution. I don't want that 2.5
to confuse you. That's just the name of
the solution, or its concentration. That's how we're
identifying it. But now we've solved
the problem. We just need to calculate this
number, so let's break out the calculator. And you get 11.8 times 2 times
1,000 divided by 65.38. And we're also going
to divide by 2.5. It's in the denominator,
it's 144. And actually our lowest-- let's
see, a couple of numbers here only have three significant
digits, so we should really only stick to
three significant digits, so it's 144 milliliters. So this right here is 144
milliliters of, we could say the solution, or of
the 2.5 molar hydrochloric acid solution. And we're done. Hopefully you found that fun.