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## Chemistry library

### Course: Chemistry library > Unit 5

Lesson 3: Limiting reagent stoichiometry- Limiting reactant and reaction yields
- Worked example: Calculating the amount of product formed from a limiting reactant
- Introduction to gravimetric analysis: Volatilization gravimetry
- Gravimetric analysis and precipitation gravimetry
- 2015 AP Chemistry free response 2a (part 1 of 2)
- 2015 AP Chemistry free response 2a (part 2/2) and b
- Limiting reagent stoichiometry

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# 2015 AP Chemistry free response 2a (part 1 of 2)

Moles of ethene produced from dehydration reaction with ethanol. From 2015 AP Chemistry free response 2a, part 1 of 2.

## Want to join the conversation?

- ugh... did I miss something? The last video I watched in the chemistry play list was Limiting Reagent Stoichiometry, when did all of this "torr", "catalyst", "K*mol", and other stuff some from? Can someone post links to vids to help out maybe?(114 votes)
- Most of the "other stuff" you're talking about is from the Gases and kinetic molecular theory section, which deals with gases. You should definitely take a look at some of the videos in the link below before trying out the questions in this video. https://www.khanacademy.org/science/chemistry/gases-and-kinetic-molecular-theory(34 votes)

- Why can you use the total volume of the gas produced, 0.0854L, as the volume of Ethene gas? Isn't that volume including the water vapor as well? Why wouldn't you need to subtract part of that volume to find out the actual volume of ethene gas in the tube?(9 votes)
- The question states the final values (volume, pressure, temperature) of the collected gas at the END of the reaction, so we must consider the problem independent of time. This means the temperature can be considered constant - as this problem would become much more complex with time-dependance and thermodynamics. As to the question about volume: we are able to use the total volume since we are using the partial pressure of ethene in our Ideal Gas Law equation. Since the temperature, pressure, and volume are assumed 'constant' the number of moles of each gas is dependent only on the partial pressure of the respective gas. If we used the total pressure of the gas collected, then the solution (to the Ideal Gas Law eq.) would give us the moles of both the ethene AND water vapor at the END of the experiment.(4 votes)

- can't we simply find the number of moles of ethanol and then use stoichiometric ratio to find moles of ethene and then find mass of ethene produced?(4 votes)
- If we assume that the reaction goes to completion, then yes, we can just use stoichiometric ratio to calculate the the amount of our products. However, since question (a) part (i) asks us to find what is actually produced, it’s safe to assume that the reaction did not go to completion. In this case, we have to use the information given in the question to figure out how much ethene is produced, which is why Sal takes the steps that he did.(4 votes)

- We have considered the pressure of vapor water but how about the volume of vapor water? Shouldn't the volume of ethene smaller than 0.0854 because some volume of the collected gas belongs to vapor water?(4 votes)
- While it’s true that the water molecules in the collection tube would take up some volume, when thinking of gasses in the “ideal” sense, we usually assume that the individual gas molecules do not take up any space in the container, since the gas molecules only make up a tiny proportion of the overall volume of the container.(2 votes)

- What's vapor pressure or partial pressure and how did we know the total pressure is the sum of the partial pressures of ethene and water?(3 votes)
- You've presumably not yet covered this on your course. There are videos on the Khan Academy related to partial pressure (https://www.khanacademy.org/science/chemistry/gases-and-kinetic-molecular-theory/ideal-gas-laws/v/partial-pressure) and vapour pressure ( https://www.khanacademy.org/science/chemistry/gases-and-kinetic-molecular-theory/ideal-gas-laws/v/vapor-pressure-example).

You will learn there that the total pressure of a mixture of ideal gases is the sum of the partial pressures. This is known as Dalton's Law of Partial Pressures.(3 votes)

- What is a "catalyst" and an "atmosphere" and "torr"?(2 votes)
- A catalyst speeds up the rate of reaction

Atmosphere and torr are two different units of pressure(4 votes)

- Were you not meant to times by 305 K in the last calculation rather than divide?(2 votes)
- It’s in the denominator so dividing is correct. Sal probably should be using a scientific calculator so he can type this all in in one step.(2 votes)

- What's metrics on the actual reaction? I don't understand all that rxn and all.(2 votes)
- Metrics could be said as the conditions and the requirements of a reaction, they are usually energy req or released, req pressure, temp, bond strength for a reaction. These aren't that useful until you come to performing actual reactions or studying Chemical thermodynamics/kinematics . It maybe used in finding % of a compound in a reactions as certain reactants can give rise to two types of compounds made of the same atom as each other (Don't ask how- it is really complicated). These are usually taught in 11th and 12th year (i.e Junior,senior,AP)

As for RXN it is usually a short form for 'Reaction', but in this case mole(rxn) means how many moles are actually formed in a reaction. (if there are any limiting reagents).

Another definition-It has units of energy per mole reaction (rxn). Per mole reaction refers to the balanced equation. For the given reaction 1 mole rxn would be the same as 1 mole of CH4(g) reacted, or 2 moles of O2(g) reacted, or 1 mole of CO2 formed, or 2 moles of H2O formed. (Source- https://ch301.cm.utexas.edu/section2.php?target=thermo/enthalpy/enthalpy-reaction.html )

Hope this helped(1 vote)

- So this question is not related to the information in the video

I checked the chemistry AP course and found it the same as the general course.

Can anyone explain to me what's happening? Is the general course AP?(2 votes) - Hello Sal

I wanna ask how did you know that the pressure of .822 atm is referring to the total pressure ? because if I tackled this problem for the first time, I would regard this pressure as the pressure of the gas as the problem says that the volume of collected "gas" is 0.0853L at .822 atm(2 votes)- this is a good question......

actually in our problem the only reactant is ethanol and it only forms the two products. so at first there is only ethanol so the pressure of gas that is the pressure of ethanol which is therefore the total pressure which is given to be 0.822 atm.

hope this helps......(1 vote)

## Video transcript

- [Voiceover] Ethene, C2H4, molar mass of 28.1 grams per mole, may be prepared by the dehydration of ethanol, C2H5OH, molar mass 46.1 grams per mole, using a solid catalyst. A setup for the lab synthesis is shown in the diagram above. The equation for the dehydration reaction is given below. So, we have the ethanol,
and then in the presence of a catalyst, we're going
to yield, after our reaction, some ethene and some
water, and then we have some metrics on the actual reaction. A student added a 0.200 gram sample of ethanol, this is ethanol, C2H5OH, to a test tube,
using the setup shown above. So this is the glass wool with ethanol, so the ethanol is right over here. There's a solid catalyst right over there. The student heated the test tube gently with a Bunsen burner until
all of the ethanol evaporated, and gas generation stopped. So, here you see the heating,
all of the ethanol here evaporates, the gas generation stops. When the reaction stopped,
the volume of gas collected was 0.0854 liters at 0.822 atmospheres and 305 Kelvin. The vapor pressure of water at 305 Kelvin is 35.7 torr. So, what you have is, in the
presence of the catalyst, you have the ethanol, it's going to react, it's a dehydration, it's
a dehydration reaction right over here, so
you're going to produce ethene and water, and
so, you cool it down, and so the ethene gets
captured at the top, it's going to be in a gaseous state. The water, it's going to be water vapor, but then, of course, once
you've cooled it down, you're going to have
liquid, you're going to have liquid water here as well. So, let's try to answer their questions. Calculate the number of moles of ethene, one, that are actually
produced in the experiment and measured in the gas collection tube, and two, that would be produced if the dehydration reaction
went to completion. Part b, calculate the percent yield of ethene in the experiment. All right, so let's tackle part one. Let's tackle part one first, and so, somehow we have to figure out the, we have to figure out the actual moles produced in the experiment and measured in the gas collection tube. So, what do they tell us here? They said until all of
the ethanol evaporated and gas generation stopped. When the reaction stopped,
the volume of gas, so this volume of gas, so
this is our volume of gas right over here, that's
where our ethene is, our volume of gas was 0.0854 liters, so we underline that, at 0.822 atmospheres, so that's the total pressure there. The temperature is 305 Kelvin, and then they give us the vapor pressure of water is 35.7, at 305 Kelvin, is 35.7 torr. The vapor pressure of water, you could view that as a partial pressure of water, and the reason
why that's useful is the partial pressure of water plus the partial pressure of the ethene is going to add up to
the total pressure here. And so, we can use that information to figure out the partial
pressure of the ethene. So, the partial pressure of the ethene. And if we know the partial
pressure of the ethene, well, then we can use the Ideal Gas Law to figure out how many moles of ethene, how many moles of ethene are
actually right over here. How can we do that? Well, just a little reminder
of the Ideal Gas Law. We have pressure, and
if we're talking about one particular thing, here we'd be talking about the partial pressure of ethene, times volume, they give us
the volume right over here, is equal to the number of
moles times the gas constant times the temperature. Well, we have the pressure,
we have-- or, we'll be able to figure it out, we have the volume, we have the gas constant. You don't have to memorize these things. Right over here, I copied and pasted what they give you at the
front of the actual AP exam, so that you don't have to
memorize these constants and things like that. So, we have the gas
constant, and then we have it in different units, depending on which one we want to use, and then, of course, the temperature, they give it to us. It is 305 Kelvin. So, if we want to figure
out the number of moles, right over here, you can divide both sides by RT, and so you could say n is equal to, is equal to R, sorry, n is equal to, we divide both sides,
if we divide both sides by RT, we're going to get PV, and I'll say the partial pressure of ethene, the partial pressure of ethene, times the volume, times the volume, over, over R times T over R times T. And so, let's figure out
everything over here. So, what is going to be the
partial pressure of the ethene? Well, the pressure of,
the partial pressure of the ethene is going to be equal to the total pressure, the total pressure minus the partial pressure, or you could say the vapor
pressure, of the water, the partial pressure of, the
partial pressure of the water. So, this is going to be equal to, our total pressure is 0.822 atmospheres, 0.822 atmospheres. Now, what's the pressure of, what's the partial pressure of the water? They tell us the vapor pressure of water at 305 Kelvin is 35.7 torr. Well, they gave us the total
pressure in atmospheres, so we have to convert
from, we have to convert from torrs to atmospheres
if we want to stay in atmospheres. So I'll do it over here. So, the pressure, the pressure
of, the partial pressure of the water is equal to 35.7 torr, which is a unit of pressure. But if we want to convert
it to atmospheres, once again, you don't have
to memorize these things. They give it to you on the test. They tell us, one atmosphere is 760 torr. So, we wanna get rid of the torr. We wanna get rid of the
torr, and we want it in terms of atmospheres,
so one atmosphere is 760 torr, so this is going to be equal to 35.7 divided by 760. So it's this is the total pressure, minus the partial pressure of the water. Let's see, the partial
pressure of the water, let me figure this out. It is going to be 35.7 torr times one atmosphere for every 760 torr, so
I'll just divide by 760, divided by 760, is equal to, and I have three significant digits here. So, 0.0470. 0.0470. So this is going to be 0.0470 atmospheres. So, let me write that down. So, the total pressure
minus the partial pressure of the water, 0.0470 atmospheres, is going to be equal to,
is going to be equal to-- So, let me just take my-- Well, I could just make that a negative, and then add it to the total pressure, so, plus, plus .822 or 0.822 is going to be equal to... So, the partial pressure of our ethene, three significant digits here, is going to be 775, 0.775. So, 0.775. And so now, we can substitute that in. We know everything to figure
out the number of moles. N is going to be equal to 0.775. The units here are atmospheres; we can always check to make sure that we get that right, atmospheres. Times our volume. Well, they tell us our volume. It's 0.0854 liters, 0.0854 liters, and then we divide that by RT. Now, which version of R do we use? Well, we're dealing with atmospheres, moles, and Kelvin, so we could use, we could use this right over here. If we had converted everything to torr, we would use this constant. So, our R is going to be 0.08206. They're giving us more significant digits. Doesn't hurt to use
them, but later we have to just realize that we're gonna convert to only three significant digits, because that's the minimum,
that's what we have. These are three, three, this
is more than three signifi-- This is four significant digits. But let's just make
sure the units work out. So, this is going to be-- So this is the units here
are liters times atmosphere, and so those will cancel
with these, divided by, we have to remember this is
in, this is in the denominator right over here, divided
by moles times Kelvin, and we multiply that
times the temperature, which is 305 Kelvin, 305 Kelvin. And we'll see, we could see that, well, this atmospheres is going to cancel with that atmospheres,
that liters is going to cancel with that liters,
this Kelvin is going to cancel with that Kelvin,
and so, in the denominator, you're dividing by moles, so that's gonna be the same thing as just
having moles as your units. And so, this is all going to be equal to, we deserve a little bit of a, we deserve a little
bit of a drumroll here. This is equal to-- So, 0.775 atmospheres times .0854 liters divided by .08206, so that gets us that, and then we're going to divide by 305. We still have that in the denominator, divided by 305, is equal to, and if we do
three significant digits, it's 0.00264. So this is approximately, I could say, 0.00264 264 moles. Did I write that right?
I have a bad memory. 0.00264, yeah, I rounded down to 264 moles. And there you have it, that's part one. The number of moles of ethene
that are actually produced in the experiment and measured
in the gas collection tube.