- 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
2015 AP Chemistry free response 2a (part 2/2) and b
Efficiency of ethanal dehydration reaction. From 2015 AP Chemistry free response 2a (part 2/2) and 2b.
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- May you tell me where did the rest of yield go ?
I mean where are the rest of ethene's moles when the reaction went to completion
- Some may have escaped the collecting apparatus, some may be unreacted ethanol, some may have gone into side reactions, etc.(11 votes)
- The video said that if the reaction goes to completion every mole of ethanol would become a mole of ethene and a mole of water. But then he said that .00434 mol of ethanol would make .00434 mole of ethene, then what about the water?(2 votes)
- According to the balanced chemical equation, there is 1:1 ratio between ethanol (C2H5OH) and water (H2O). That means that if the reaction occurs, and 1 mole of ethanol is used up, then 1 mole of water is produced. Or, if 7 moles of ethanol are used up, then 7 moles of water are produced. OR, to answer your question, if 0.00434 moles of ethanol are used up, then 0.00434 moles of ethanol are produced...(2 votes)
- Is the question from part B similar to the percentage error question?(1 vote)
- In part 1, the answer is based on when all of the ethanol had evaporated, so the reaction stopped. Why does this not imply that the reaction "went to completion"? In other words, what physical part of the reaction isn't complete in part 1, that causes the yield to be different in the two parts?(1 vote)
- In questions like this, how do we know that we are to use gas laws?(1 vote)
- How else could you solve it? If you’re the given volume, pressure and temperature of a gas in the question that’s a pretty big hint.(1 vote)
- So to calculate the percent yield, you divide the actual yield by the theoretical yield and then multiply by 100 % right?(1 vote)
- I did not understand the term gas constant mentioned in2015 AP Chemistry free response 2a part 1 of 2(1 vote)
- Have you watched the videos on the ideal gas law? That would be a good place to start.
- Can someone tell me what those things are infront of the chemical equation? the one with the delta symbol(1 vote)
- ΔH°298 is the enthalpy change of the reaction. Essentially it tells us how much heat is being exchanged between the reaction and the surroundings. Being positive it is an endothermic reaction which means heat is being absorbed by the reaction from the surroundings. The delta symbol means change in math, the degree symbol means this number is taken at standard conditions, and the 298 means the temperature this happens at is 298 K.
ΔS°298 is the change of entropy of the reaction. Essentially how spread out does the matter become over the course of the reaction. A Positive value means the matter becomes more disperse compared to when it started. This makes sense since the products are two molecules compared to the one molecule on the reactant side. The other symbols mean the same as the enthalpy explanation.
Hope that helps.(1 vote)
- Part A
Determine the balanced chemical equation for this reaction.
Enter the coefficients for each compound in order, separated by commas. For example, 1,2,3,4 would indicate one mole of C8H18, two moles of O2, three moles of CO2, and four moles of H2O.(0 votes)
- The balance equation coefficients are 1,17,8,18.(0 votes)
- [Voiceover] Alright I'll tackle. In the last video we did the first part of part A, now, so the second part of part A. So, the second part of part A, they say, "Calculate the number of moles of ethene that would be produced if the dehydration reaction went to completion". Well, this is the dehydration reaction right over here, and they're telling us, they're telling us that we start with 0.200 grams of ethanol, so, if we start with 0.200 grams of ethanol, and if we figure out how many moles of ethanol that is, well, for every mole of ethanol, if we have the reaction go to, if the dehydration reaction goes to completion, for every mole of ethanol that we start with, we're going to have a mole of ethene, and a mole of water. So, if we can just figure out how many moles of ethanol this is, then we'll say, "Okay, if this were completely react, we would have that many moles of ethene". So, let's figure this out. So if we say, so, we have ethanol, so for ethanol, we are starting with zero .200 grams, and we wanna convert this to moles, so we wanna multiply this times, we want grams in the denominator to cancel out with this grams, and moles in the numerator. So, one mole of ethanol is, as a mass of how many grams, well, they tell us that earlier on the problem, they say, "Ethanol, molar mass 46.1 grams per mol", so, ethanol, molar mass of 46.1 grams per mol, or another way to thinking about it, one mole would have mass of 46.1 grams of 46.1 grams, and so if we do this, we are going to get 0.200 over 46.1, and then grams cancel with the grams, and that going to be how many moles we have, and this is going to be equal to, we have three significant digits that we're gonna be with three significant digits divided by three significant digits, so this is going to be all right, so we clear this, so .2, I could write 00, but it's going to, for the calculator's purposes this is the same thing, divided by 46.1 is equal to 0.0043, I want three significant figures here, so 43 I'm gonna round up, 434, .00434 so 0.00434 moles. So, we've seen so far, if the ethanol were (mumbling) the reaction, if we have this many moles of ethanol, well, and if they completely react, well, then we should end up with that many moles of ethene, and so if we have, if it we get a dehydration, I'm gonna write it this way, if dehydration dehydration reaction goes to completion goes to completion every every mole of ethanol would be converted to a mole of ethene and a mole of water. I shouldn't write the shorthand there, a mole, if I'm writing it out, mole of and a mole water. So 0.00434 mole of ethanol would yield, that same number of moles, 0.00434 moles of ethene. So, if the reaction went to completion that how many moles it would produce. What we actually measure is a smaller number than that, so the reaction didn't go fully to completion. So now let's tackle part B, let's tackle part B, and they ask us, "Calculate he percent yield of ethene in the experiment". Well, the percent yield is going to be how much we got, divide by how much we would ideally have gotten if the reaction went fully to completion. So yield yield is going to be equal to how much we actually got, which we figured out in part one, so 0.00264 moles over what we actually got, i'm sorry, what we actually got over what we'd have ideally gotten if the reaction went to completion, so divided by 0.00434 moles, and this is going to be equal to this is going to be equal to so, 00264 divided by 00434, is equal to, and let's see, we have three significant digits here, so 60, I can say .608, or I could say 60.8% yield. So, I'm gonna write this as actually this is gonna, I'm rounding, so I can say 60.8% yield, and there you go.