Tip:
Highlight text to annotate it
X
Hi everyone, this is your first video for Unit 7. The topic is stoichiometry. Stoichiometry
is a type of calculation that is based in a balance chemical equation. This will correspond
to chapter 11 of your text. Let's go ahead and get started. As I said before, stoichiometry
is based on the quantities that we know using a balanced chemical equation. So as we're
working through these stoichiometry equations, it will be very important that we take note
of and check off a few things on our list as we're working through these. First of all,
for all stoichiometry equations you'll need to have a balanced chemical equation. You
will write that out and balance it if it's not already given to you. We will use the
coefficients to determine a mole to mole ratio. Think back to your hydrogen gas lab that you
did last unit. You used a mole to mole ratio to convert from moles of magnesium to moles
of hydrogen. Then lastly, it's going to be even more important than before to make sure
that you are accurately labeling all units within the dimensional analysis problems.
We're going to be dealing with more than one substance within a dimensional analysis equation
now, so labeling within each of those little railroad steps will help you keep track of
which substance you're dealing with and which unit of measurement you're dealing with as
well. The rest of this video will consist of working through two practice problems.
These practice problems correspond to two different practice problems on your worksheet,
so you'll want to make sure you have your Stoichiometry Practice 1 worksheet handy as
we go through these. That way you have a couple you've worked through as you try to finish
the rest. The first problem corresponds to number 1 of your Stoichiometry Practice Sheet
1. The equation is given to you. Nitrogen gas combines with hydrogen gas to form ammonia,
which is NH3. You'll notice that it's already balanced. I have nothing written in front
of my N2. There are three units of hydrogen and 2 units of ammonia. The "nothing" in front
of the N2 is actually a one. Remember if we don't need to add a coefficient, we just assume
that there's a one in front. So what this reaction means is that one mole (because the
coefficients tell us how many moles are involved in a reaction), so one mole of nitrogen gas
combines with 3 moles of hydrogen to produce 2 moles of ammonia gas. This gives us a mole
to mole ratio of 1:3:2. And so that's what we'll use to complete the rest of this problem.
The question is asking, "How many moles of ammonia are produced when .5 moles of nitrogen
gas reacts with hydrogen. So this is a very simple reaction because I'm already given
my units in moles. So I'll start with my .5 moles of nitrogen. And this is what I mean
by the importance of labeling fully, because my mole ratio now is the next step. In the
mole ratio, I need the moles of nitrogen because that's what I have here, and moles of ammonia,
because that's what it's asking me to solve for. So like with all other set-ups in dimensional
analysis, one mole of nitrogen from my equation will go on the bottom, then moles of N2 cancel.
On top will be the mole value for the substance that I'm changing to, which is the ammonia.
Two moles of ammonia, NH3. Go ahead and wrap this calculation up. The 0.5 times 2 gives
me one, divided by one, because we multiply everything together in the numerator. Divide
by everything in the denominator. So this gives me one mole of NH3 (one mole of ammonia).
Basically these tells me that if I start my reaction with 0.5 moles of nitrogen, because
of my mole to mole ratio between the hydrogen and the ammonia, I'll produce one mole of
ammonia when I start my reaction with 0.5 moles of nitrogen gas. The next problem: Clacium
chloride and water react the produce calcium oxide and hydrochloric acid. This will correspond
to #2 of your worksheet. Note I do not give your a balanced chemical equation, so go ahead
and pause the video right now. Use your information from the top of the page to balance your equation.
Go ahead and resume the video when you think you have your equation correctly written and
balanced. Okay hopefully you're double-checking your written and balance chemical equation,
and you should hopefully have it correct. If you do, you'll notice that one more of
CaCl2 (that's your calcium chloride) combine with one mole of water to produce 1 mole of
calcium oxide and 2 moles of hydrochloric acid. So we can check off that the equation
is written and balanced and move down into the actual chemical equation. This is asking
how many grams of calcium that are produced (so this, right here). When I start my reaction
with 4.0 grams of calcium chloride, which is this reactance right here. I quick roadmap
for this. Because we are starting measuring in grams of calcium chloride, you should remember
from mole calculations that you cannot just go from grams to grams. You need to go from
grams of calcium chloride into moles of calcium chloride. From there we can use our mole ratio,
which I'll abbreviate with a capital "MR" change 2 moles of calcium oxide. Once we have
moles of calcium oxide, we can they change to grams of CaO. So just like with previous
calculations, we need to change whatever unit we're starting with to moles (if it's not
already there), and then from moles you make that switch from a different measurement or
a different substance. And that's were the whole stoichiometry comes in. We're using
those mole ratios to switch from one substance to another within a balanced chemical equation.
So my starting point is 4.0 grams of calcium chloride. Make a nice, long bracket because
I will use it. So I need to switch to moles, so I need to know my formula mass for calcium
chloride. Which happens to be 110.98 grams. Put that on the bottom. CaCl2 to one mole
CaCl2. Looking right away, noting that we cancel grams of calcium chloride, and now
adding moles of calcium chloride. Next step is the ratio. So according to my balanced
chemical equation, I have one mole of CaCl2, and that is relative to one mole of calcium
oxide taken from my balanced equation. Notice moles of calcium chloride will cancel. This
is why we call it dimensional analysis, because you are using the units, the dimensions of
measure, to set and determine where everything goes within your bracket. Next step: So let's
revisit our roadmap right up above here real quick. We have taken care of grams of calcium
chloride, we have care of our moles of calcium chloride, we have used the mole ratio, which
was right there. The next step is to take my moles of calcium oxide, which I am now
labeling, and convert that to grams of calcium oxide. So grams of calcium oxide in one mole
of calcium oxide. The formula mass is 56.08 grams. Moles of calcium oxide will cancel.
I'm left in grams of CaO. And that is the unit I'm looking to measure for, so go ahead
and circle that if that helps you to identify your final unit, and you know that it's your
final unit because when you look up the initial problem, it asks you for grams of CaO. Once
you had that in numerator you are done. You can cancel everything else out, you have finished
your problem. No it's just a matter of some simple math. Multiply everything through the
numerator. You should see that you have a total of 224.32 divided by everything in the
denominator. 110.98 gives me a total of 2.02 (three sig figs because we started with 3).
Grams of Ca0 produced when I start my reaction with 4.00 grams of CaCl2. Good luck as you
work through these. Remember everything starts with a balanced chemical equation, and the
labels determine what order you proceed through the problem in. So make sure that you're taking
time to pay close attention to those labels. Good luck. Let me know if you have any questions.
We'll work more on this in class.