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Rcteronces]This question has multiple parts. Work all the parts to get the most points.DDT; an insecticide harmful to fish, birds, and humans, is produced by the fo...

Question

Rcteronces]This question has multiple parts. Work all the parts to get the most points.DDT; an insecticide harmful to fish, birds, and humans, is produced by the following reaction: 2CoHsCl + Cz HOCls C1HOls + HzO GhlorobaTtana chloral DDTIn & government lab, 1168 g of chlorobenzene is reacted with 473 8 ofe chloral:What mass of DDT is formed, assuming 100% yield?MassSubmitSubmlt AnswerTry Another VerslonItem attempts remalning

Rcteronces] This question has multiple parts. Work all the parts to get the most points. DDT; an insecticide harmful to fish, birds, and humans, is produced by the following reaction: 2CoHsCl + Cz HOCls C1HOls + HzO GhlorobaTtana chloral DDT In & government lab, 1168 g of chlorobenzene is reacted with 473 8 ofe chloral: What mass of DDT is formed, assuming 100% yield? Mass Submit Submlt Answer Try Another Verslon Item attempts remalning



Answers

DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: $$ 2 \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{HOCl}_{3} \longrightarrow \mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}+\mathrm{H}_{2} \mathrm{O} $$ $\begin{array}{lll}\text { chlorobenzene } & \text { chloral } & \text { DDT }\end{array}$ In a government lab, $1142 \mathrm{~g}$ of chlorobenzene is reacted with $485 \mathrm{~g}$ of chloral. a. What mass of DDT is formed, assuming $100 \%$ yield? b. Which reactant is limiting? Which is in excess? c. What mass of the excess reactant is left over? d. If the actual yield of DDT is $200.0 \mathrm{~g}$, what is the percent yield?

This question is going to take a look at striking geometry and determining limiting re agents and the overall yield or percent yield of a product. So in this question were given this balanced reaction between chloral benzine and chloral to make DDT. And we're giving these starting masses of each of the two re agents. And as to use that to determine how much product we can essentially make. So first we need to determine the limiting reagent. So let's start with what we're given 1142 g first to the chloral benzine. And we need to relate this to moles because that's what we always do in any strike geometric problem. Um masses sort of meaningless in terms of those re agents. So moles is much more useful here and necessary. So the molar mass of the chloral benzine, which I'm just abbreviated right here is ClB that's the chloral benzine, 112.56 g per mole. And the balanced reaction tells me that it requires two moles of chloral benzine To produce one mole of the DDT. The final step in the process then Is to take then the molar mass of the DDT, which is 354.48 g per mole and multiply it by the number of moles of DDT produced. And that will tell us then the mass of DDT that can be produced from this reaction. We do the same thing with the other re agent. The chloral here molar mass of the chloral. Again, we'll get by adding up each of the individual Um atomic weights in the formula to get the formula masses 147 38 grams per mole. And the strike geometric ratio tells us for every one mole of chloral that we react. We're going to produce one mole of DDT. And again the final step in this process is to determine based on that information and the molar mass of DDT which of these two re agents is the limiting reagent. And therefore that defines for us the maximum mass or quantity of DDT that we can produce. So let's go crunch some numbers now. So in crunching out some numbers here we determined that overall this was the most significant thing. What we really need to find out when we determine the limiting reagent is how many moles of products are produced and um whichever re agent produces the fewest number of moles of product. Then we wanna multiply that number by the molar mass of the DDT to determine the maximum mass that can be produced. And when we crunched the numbers, we find that in the case of the chloral benzine we can produce 5.073 moles of DDT. But only 3.291 or 23 sig figs 3.29 moles of DDT based on that given mass of the chloral. So the chloral then is my limiting region. I'll just put an L r. By buy it. And the chloral Benzine then would be my excess re agent. And based on this given number of moles of DDT. I multiplied by the molar mass. I end up getting the mass produced or the maximum mass that can produce the DT. Based on this limiting reagent to three significance is 1170g. then of the DDT. Yeah. Now we're also asked to take this mass. Then this is the theoretical yield. And I'm just going to scroll up here for a minute and compare it to the actual the actual yield of The DDT which we were given in the question as 200 g. This is my actual yield divide that then by the theoretical yield which is The 1170 grounds. And that should give us then uh the percent yield of this product. So let's do some more number crunching. All right. So in doing and dividing this problem out, We determined here then that 200 divided by 1170. Give us a fraction of .171 To express that as a percentage will multiply by 100. And overall the% 17.1%. The final question part of the question asked us, hey, how much accessory agent are we going to have left? Well, we know the excess re agent up here is the chloral benzine. And so again, let's go back to what we know. In terms of the DDT product, we know that based on the limiting re agent, the maximum amount of DDT that we could produce in terms of moles. Is that 3.29 moles. And based on the balanced reaction, We know that it takes two moles of the chloral benzine to produce one mole of DDT. So let's use that racial here. And finally let's multiply then that many moles of chloral benzine that reacted to produce the DDT by its smaller mass. And when we do that, crunch the numbers, we determine that What actually reacted was 741 grounds of this actually should say chloral benzine nut DDT. This is the amount of chloral benzine that reacted to produce the DDT. And so how much regent was left over? Well, how much did we have to begin with? We had 1100 42 grounds From that, 741 of those grams reacted. Therefore when we do the math here, crunch the numbers out. Just a little over 400 or more specifically 400 one g of the chloral benzine is like unredacted or left behind or in excess. Based on this problem

All right, so we have to find out what defend the reaction to synthesized DDT from Corbyn's in and Crow. So were the found of massively TPM produce their mean emitting radiation, which, when you access massive excess creation and also the best Thank you. So you know that the my answer part A We have to find a name for every Asian. So we have to self apart be, you know, the tense off apart. A. So first off, I would like to come Veronica Mass to number most. So I already convert that over here, which is that you take our mass divide by number mass. Okay, second thing we could. Your father? Which one? The limited way agent. Okay. And so we're going to food the way we at riff our Ah ah, Karami thing. What would be the requirement? Limbo most off our Ah, coral. Okay, then. More than that, the motorway show ihsaa 2 to 1. So the corresponding ah require limp. Um would be equal to 10.16 The number mo off cope and center Vibert. See why the bite to So you're 5.8 to be the lumber most Okay, so I mean, you can see that if you want it for them. We at with Paula Corbin. See the require Months of Corvo is way higher than the one that we have. So therefore, the corporal will be the limiting we agents a reward if, ah so for party. And also our car Corban soon will be in excess. So we are here. So for that. Okay, so the corresponding number more off our T t t will be equals. Teoh were using call wrote to to to do a collision. The motivation is 1 to 1. So you could see the 3.96 more and you would have a corresponding mass. Ah, the mascot recorded by taking the lumber most tandem or the mass. So we have for 14 carbon ni um hydrogen. And then we have ah five Corinne. So you're fived. I'm 35.45 and every we have corresponding amasses 11 65 gram or around 1170 gram And this will be the mass off DDT being pro deals. So we are here, so for the mass. Okay, So what were massive excess? We Asian the excess waging. All right, so Ah, we we know that. Ah, Cold and Cindy's excess. So therefore, um, the lumber the blow off. Um Carro Band scene Coral band scene Ah, we meaning at the end should be close to the total number mo 10.16 subtracted lumber Modi just Teoh we act with ah that way of coral. So the lumber movie recover we? Because to the limbo, more corporal times to be got 1/2 a 1 to 2 way show. All right, so we're 10 Porn 16 sub track two times three foreign to through newly. And then we should have Ah! 3.58 mo And the number of mass Ah, the mass off a cold and say we may will be equal stiffy foreign fighters times to move That massacre been seen and it will We will have a fall two point Ah, six grand We remain so we already fund of massive excess. Okay, So if the production ah, the actual years 200 gram or did it evolve would be the percent issue? Oh, here. So from out clearly go. You were someone 2%. We have ah 11170 gram. And for the percentage you so is party yes, equals to our actual you divide by the percentage after Ah, the vertical you. And then I was like, I want your percent and we will have Ah 17 palm, one percent as 2% you

Okay, So for this question, they're saying 1142 grams of clo benzene is reacted with 485 grams of claw to form Dutt D d t. Right. So first, we must write out D balance equation. So this is clonal benzene plus clear Dios DDT, which is C 14 h nine c l five plus H 20 so this is the balance equation. So the first part that asking for what mass of DDT is form and we're assuming 100% yield. So the first thing we have to do is convert the grams of clo benzene and the grabs of chloral into Mo's to determine what is he living re agent. Right. So let's do DD TIF. Let's Duke low benzene first. So I have 1142 grams of chloral benzene, right and ivory calculated the molecular weights of low benzene, and it's 112.56 grams. And that's perm. Oh, all right, So if you want to calculate molecular mass by yourself, all you gotta do is consider how many moles of carbon we have multiplied by D a. Moeller massive carbon promo. And then you do that for hydrogen and chlorine, and you should get around the same number Opinion review around or not. Now I got template 15 Most of kuo benzene of C six h five C L Now, for the next one full core, we have 485 grams. Uh, the molar masses 147.38 grams. And that is Paramo. Put that into D calculator and weak gets 3.29 Mose of Claro So C two h o c l three Now, uh, I guess I'll use clo benzene. It doesn't matter which one you pick. So we have 10.15 most of chloral benzene, and I just look at the multiple ratio we see that's a 2 to 1 ratio. So to most of to most of quo, benzene is needed to react with one mo off tomorrow. But we put that into the calculator and we should get five point 5.75 mose of, uh, quo. So we see that we only have 3.29 most. Of course, this means that this is the limit free agent. All right, So, uh, I write this in a different color. So this is the limiting reagent. Now, what we do now is we use a living every agent to calculate how much product we have because the amount of product form is limited by delivering the region. Right. So we have 3.29 three points to nine those of koro. So I think I would see to h o C. L three. Now, let's look at the multiple ratio of Claro and uh, T d t You see that it's 1 to 1 ratio, So one more of C two h o c l three forms one mo of Ah, I'm just really like DVT, right? And I think they're asking for how much What is the mass? Okay, so one mo of D d. T have already calculated the molar mass, and it's 300 54.49 grams. Put that into the calculator. We get in 1166.3 grams of DDT. So that is the answer to part A. Now, for part B, they're asking for Which is the living? Uh, which reacting is the limiting one, right? So we've already determined what the live anyone is. Uh put a here now for B. We've already determined the women re agent as desk lower. Right? So that is the answer for the first part of part B. So, for Oh, is he living region? And then I think they're asking for how much which is leaving region, which is an access. Okay, so which is an excess, so obviously don't want that isn't dealing me. Region is an excess, and that is Ah, chloral benzene. So this is clo is a liming re agent and then chloral benzene. These include mincing. Yeah, it's close. Benzinger is the excess now for part C that asking what mass of the excess reacting is left over. So all we gotta do is calculate how maney mose of, uh C six h five c l is left over and then converting most into converging most into mass. Right? Okay, so we have 3.29 grams. So 3.23 point 29 Most of clear. Oh, the multiple ratio of chloral and quarrel Benzene is 1 to 2. Right? So one meow of quarrel needs to most of for benzene one meow of I know that zit. Yes, so we put this into D calculator and we get 6.58 most. So that's how much chloral benzene we need to use. Or that that's how much has been reacted. And from here we have template 15 miles. So you have Temple 15 moles of cool benzene. We use 6.5 because that's how much is needed to react with 3.29 miles of glow, and we get 3.57 most. So that's how much clo benzene we have left. Then we used the most of clo benzene we have left, and we convert that into Graham. So one mole of CO benzene is equal to 112.56 grams. Put that into the cockpit and we have 401 points. Eight grams of cool benzene left now for Part D, they're saying if the actual yield of DDT is 200 grams, what is D percent yield? Right. So let's just go up. We see that the theoretical yield is 1166.3 grams right? And the actual yield that give us is 200 grams. So let me write the this is the actual and in 1166.3 grams is the deal red ago. So the formula we need to use is actual yield over deal. Radical yield times by 100 is equal to deep percent yield. Right, So let's plug in values. So actually, yield is 200 over theoretical, which is 1166.3 times by ah 100. And we should get about 17% as our answer. So this is the answer for party. This is for part C. Uh, this is for B and then we have this as the answer for part a.

This question asks us to find, ah, the great constant K for a particular first order reaction. And it's in regards to DDT, which is apparently a pesticide. And then the second part of the question asks us to find how long it takes to go for one concentration to another. So we'll use our integrated rate laws for that. So being that this is a first sort of reaction, I just go to my table and find that the half life of a particular first Order reaction is a ratio of 10.693 over K. So knowing that the half life is 56 days, I'm just gonna go ahead and do a little bit of algebra here will solve for unknown Kay. Here, 56 times k equals Wait 693 When we divide both sides by 56 I get K equals zero point 012 375 and that's per day. Okay, so that's gonna be over. I'll just put day, There's my K. Okay, We could switch over two seconds, but that would probably be a little bit too much work. We're going to get a pretty large number and day. So we'll just keep this in days now if we go over to, um, our second part here. So we have our K. Now we have enough to figure out what our time is. How much time does it take to go from one concentration down to a smaller or more diluted concentration? We do have our original concentration given in the problem and our final All right, those values down. We also know rk as well. So we're just finding our missing tea in this particular problem, and we'll figure out how long it takes to go from one concentration to another. So I went ahead and for shortcutting purposes, I just went ahead and subtracted out. You know, Ellen of our original concentration, That's where a subzero. Okay, so we'll have Ellen of a minus ln of a sub zero, and those values will give. When I subtract those out, I get negative six point or worry. So in my calculator, I just did. Ellen, uh, the Ellen of A is one point for one times tend to this negative seven power, and I just subtracted out the original ln of 8.75 times 10 to the negative by. That's a little bit more concentrated. So that will give me if you just put down your calculator who hit negative 6.43 So we get negative 6.43 on the left, and that equals our great constant, which is put a negative sign out there. 0.123 75 times tea. Now, when you divide both sides by the negative 0.12375 you end up with T equals I get along decimal, I get 5 19.6472 But because it's 5 19.6 I rounded it up to 520 days, so that would be your final answer. It takes 520 days to go from your original concentration, which was the 8.75 times 10 to the negative five, down to the lower concentration of 1.41 times 10 to the negative seven. And we used what's called our integrated rate law, too. And the information given to find time


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