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Athene c2h4 molar mass of 28.1 grams per mole may be prepared by the dehydration of ethanol c2h5oh molar mass forty six point one grams per mole using a solid catalyst a set up for the lab synthesis is shown in the diagram above the equation for the dehydration reaction issue 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 zero point two zero zero gram sample of of ethene of ethanol this is the 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 0.085 four liters at 0.8 to two atmospheres and 305 Kelvin the vapor pressure of water at 305 Kelvin is thirty five point seven Torr so you have is the you 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 it's once you cool it down you're going to have liquid you're going to have liquid water here as well so let's let's try to answer their questions calculate the number of moles of ethene one that are act produced in the experiment and measured in the gut in the gas collection tube and - 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 1 let's tackle part one first and so somehow we have to figure out the we have to figure out the actual moles producing the experiment and measured in the gas collection tube so what do they tell us here they said until all of the ethanol evaporating 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 0.08 5 4 liters so let me underline that at 0.8 - 2 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 point 7 at 305 Kelvin it's 35.7 Torr so the vapor pressure of water you could view that as the 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 could 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 we're 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 can be able to figure it out we have the volume we have the gas constant you don't have to memorize these things that 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 the gas constant and we have it in different units depending 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 egos equal to we divide both sides if we divide both sides by RT we're going to get P V 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 zero point eight two two atmospheres zero point eight two two 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 thirty five point seven Torr well they gave us the total pressure in atmospheres so we have to convert from we have to convert from Tour's two 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 thirty five point seven tor which is a unit of pressure but if we want to convert it to atmospheres once again you'd have to memorize these things they give it to you on the test they tell us one atmosphere is 760 Torr so we want to get rid of the tour we want to get rid of the tour and we want it in terms of atmospheres so one atmosphere is 760 Torr so this is going to be equal to thirty five point seven divided by seven hundred and sixty so it's this is a 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 thirty five point seven tor times one atmosphere for every six hundred 760 Torr so I'll just divide by 760 divided by 760 is equal to and I have three significant digits here so zero point zero four seven zero zero point zero four seven zero so this is going to be zero point zero four seven zero atmospheres so let me write that down so partial the total pressure minus the partial pressure of the water zero point zero four seven zero 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 point eight to two or zero point eight to two is going to be equal to so the partial pressure of our ethene three significant digits here is going to be seven seven five zero point seven seven five so zero point seven seven five 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 zero point seven seven five the unit's here are atmospheres we can always check to make sure that we get that right atmospheres times our volume well they tell us our value it's zero point zero eight five four liters 0.08 five four liters and then we divide that by RT now which version of our 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 were converted everything to tour would use this this constant but so RR is going to be 0.08206 they're giving us more significant digits it doesn't hurt to use them but later we have to just realize that our that we're going to convert to only three significant digits because that's that's the minimum that's what we have these are three three this is more than three significant 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 count will cancel with these divided by we have to remember this is in this is in the denominator right over here divided by Mol's 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 atmosphere as that leader is going to cancel with that leaders this Kelvin is going to cancel with that Kelvin and so you're in the denominator you're dividing by moles so that's not going to 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 zero zero point seven seven five atmospheres times point zero eight five four liters divided by point zero eight two zero six so that gets us that and then we're going divided by 305 we still have that in the denominator divided by 305 is equal to and if we do three significant digits it's zero point zero zero two six four so this is approximately I could say zero point zero zero two six four two six four moles and I write that right I have a bad memory zero point zero zero two six for you I rounded down to two six four moles and there you have it that's part one the ACT the number of moles of ethane that actually producing the experiment and measured in the gas collection tube

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