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Synthesis of Methyl Benzoate Theoretical Yield and charthelp5 grams sodium hydroxide, 50 mL distilled water and 40 mLethanol added to a beaker and heated5.4 distill...

Question

Synthesis of Methyl Benzoate Theoretical Yield and charthelp5 grams sodium hydroxide, 50 mL distilled water and 40 mLethanol added to a beaker and heated5.4 distilled benzaldehyde and 1.5 g acetone were mixed andadded to the mixture above to be heated.product: 52.009 g

Synthesis of Methyl Benzoate Theoretical Yield and chart help 5 grams sodium hydroxide, 50 mL distilled water and 40 mL ethanol added to a beaker and heated 5.4 distilled benzaldehyde and 1.5 g acetone were mixed and added to the mixture above to be heated. product: 52.009 g



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A mixture of benzaldehyde and formaldehyde on heating with aqueous $\mathrm{NaOH}$ solution gives (a) Benzyl alcohol and sodium formate (b) sodium benzoate and methyl alcohol (c) sodium benzoate and sodium formate (d) benzyl alcohol and methyl alcohol.

So we're giving the reaction. Ah, we have different to ration. The first reaction is deceiving yourself, Doug. If our either. So we have two f new mock You going to react with each other to form my evil beaver show over here. And then also, there's a side reaction. They will be sure Ana reaction to the offender will undergo Ah, dehydration to form. Ah, after me. All right. So if we ah, have 50 grand Avenue to start with and they put your for your five point ah, nine Graham off fever, what would be the ah percentage you What will be the percentage you? So the percent it's you, its Sequels to our, um um the experimental. Or you can see the match experimental. Um, you all further for radical, real free, radical you. So it's the foetal. What would be the if one of your first single? Well, what we're there for vertical. You tires 100% and then we can find out. Ah, percentage. You Okay, so for this case Ah, let's go look at our equation for two. You know it will form one. Ah, either to f inal form one either. So We're going to cover the theoretical you first fee. Ready, Cole, you Is this so we know that the mood away show off. The reaction is to afternoon, too. One either ratio. So we should have to found the lumber Morse off afternoon. So I just e t o h s f no. So we have 50 Graham or effort? No, to start with. And the afternoon. Oh, ah, for Miller a see Ashley ch two away. So we have to find them on the mass. So, um so that's father. Will the mass first Super 16 1 It would be cause for six grand per mo. All right, Sylvia. 46 gram for mo so we can find on Bumble, so we'll be equals to 1.1 mole. Well, definitely. Okay, so we go back to our location again. We have to afternoon the motorway Shit between the after no and ive. There will be 2 to 1. You can say from the reaction equation. So we want to found Ah, Either the limbo more would you for reveals the 1.1 d y bites you because 2 to 1 ratio. So when from one day of by two. It would really couldn't see well, for 55 more. All right, so we have to convert that. Ah, assumed this reaction going to wonder if Assange complete. So is we have 7.55 more theoretical. So what would be the mass off? Ah, you for produce. So we are going Thio Mother plan our morning mass of fever with the number most so were sealed for 55 most times the lumber Moore's off fever. Okay, so that's coming here that I'm Baltimore UV B is a pretty big mark. You Ah, for one. Uh, actually, we have to Ah, afternoon, Come buying referred shoulder. But we have a water spinning nose and the process, so we know that, um, they're total on the mass Off Avenue is 46. So we just went. Bye bye. Forces buying to be way up to afternoon. Ah, educator. But these loose of water So minus 18 the water. So we know that we recently four grand for more form with a mass off that ether and everyone by by the humble most we should be able to find a 40.7 grandfather of theoretical. You All right. So the looks that we go back, we're going to plug our free radical you over here. So it's 44 41 7 So 40 plus if incorrect. So we have 35.9 d white by 40.7 time. 100%. It will be equals to 88 0.2 percent will be the percentage you. Okay, So, poppy, if 45% of the offender doesn't pull the receiver So what would be the Master Evan introduce? Because 45% off them assume for 5%. It doesn't put to use their either than they must go to the side reaction. So, um, we are going thio. Okay, What would be the weight off that afternoon first, So Ah, let's go back where? 45%. So we have 50 grand wealth afternoon to start riff. So your mother piper so points for five or 45%. So it will be 22.2. Ah, for 22.5 gram of avenue. Okay, so, um, we want you know, how many limbo more f inal from here. So it would be 22.5 year by 46. Ah, we will have several points for eight night. Most off. Ah, you know. Okay, so we go back to our side reactions. So for each after New Year, four years, one evidence. So it's 1 to 1 ratio. So the number of most off our ah ah ch to siege to we also you co CEO Seo Point for 89 most All right, so we just need to call Write that to the mass. So we just leave Thio, um, take our love almost mummified by the more than mass over time Some of the mass What would be the more than mass again? We can't just called me that. Ah, it will be equals to ah, 28. So a 28 grand for mole, Not just missing units a safe time. So you breathe 13 0.7 graham off our ch to say she'll produce if we have four or 5% f inal doesn't produce their evil on desert. The answer of poppy

So the limiting reagent in a chemical reaction is defined as the reactant substance which is consumed completely at the completion of the reaction and the yield of our products will depend on the amount of limiting reagent we have. So we have the following equation c. two H 5 H. As our liquid, add 302 as our gas. That gives us two Co two as a gas And three H 20 as a liquid. Where we have the mass of one mole of ethanol is 46.7 g. The mass of three moles of oxygen Is 96 g. So because the massive ethanol is less than that of oxygen, we can see that are limiting reagent is ethanol so we can determine the massive ethanol via the density. And so what we have is 9.789 g per mil, multiplied by 462 mil To get 3.65 g. So in this reaction we have one mole of ethanol that gives us three moles of water. And so this tells us that if we have 46.07 g of ethanol, this generates three moles of water. So the number of moles of water that will be produced by 3.65 g of ethanol. We can determine that. So we take three moles divided by 46.07, Multiplied by 3.65. We got not .238 moles of water. And then we can determine the mass of water. Using the moles and the molecular mass from the periodic table. Multiply the moles by 18 g per mole. Councils are are moles and we get 4.28 g. So finally, we have the experimental yield that is the yield of water collected. That has given us 3.72 mil, which is equal to 3.72g. If we consider the density of water, so a percentage yield, It's equal to the experimental yield 3.72 grounds, divided by the Theoretical yield 4.28 g. Multiplied by 100, we get 86.92%.

Okay, We've been given the freezing point change. So let's go ahead and use that to find our morality. Thanks. It's 5.5° RK. F for water is 186. So, our morality is going to come out to be 3.0. Okay. And thats moles of solute per kilogram of solvent. Well, I know I have 100 g or .1 kilograms. So I've got .30 moles here. Mm. So now I've got two different substances. So we're gonna need to work in terms of variables. So I'm gonna make X equal to the moles of my nothing. All Okay, so that means that .3 minus X will equal Are moles of the purple alcohol. Mhm. And I know that the grams of my methanol plus the grams of my proposal alcohol have to equal the total grams which was given to us As 14.2 g. Mhm. So I'm going to say that I have X moles of nothing All and I'm going to change malls two g By multiplying by the Molar Mass, which is 3242. Mhm. And I'm going to add the 0.3 zero minus X. Malls. We're gonna go and change moles to grams there with that proposal alcohol Which has a molar mass of 60.95. And that whole thing has equal 14.2 g. Mhm. So this bit right here is just the grams of the methanol and this is the grams of the propranolol. So really all we need to do is just some algebra and solve for X. Okay, so I'm just simplifying all of this. Okay, When I get 3.83, It was 28 .053 x. So X is going to come out here to be .14 moles. And I called X methanol. So that's my moles of methanol and they want a mole percentage. Okay, so more percentage, So that's going to be the moles of the methanol .14 divided by our total moles. We got up here 30 times 100 And we'll get that. It is 47 methanol.

For this question were given a mixture that contains 50.0 grams of ethanol, 50.0 grams of water and 5.0 grams of sugar, and were asked to calculate the percent ah mass of each compound in the mixture. So that would be P percent massive, the ethanol to be percent mass of the water and p percent mass of the sugar. And we're also asked, how many grams of a mixture um would we need in order to have 1.5 grams of sugar? And how many grams of the mixture should he have to have 10 grams of ethanol? So for the first part of the question, you're asked to Seoul 4% mass of each compound. Now percent mass is equal. Teoh the mass Um um be compound that you are trying to find the percent mass of. So, let's say percent mass. Um, X is equal to the mass of X, divided by the total mass of the solution, and that is multiplied by 100 to get a percentage. So that is the formula we're going to be using throughout this problem. So first step is we are going to find the percent mass of ethanol will start with ethanol because that's what was given first. So the percent mass of ethanol I'm just gonna rate e t h for sure is equal to the mass of ethanol, which is given to be 50.0 grams, and that will be divided by the mass of the total solution. So to find that we just have to add up the three masses be massive path and all the water and the sugar. So that gives us 105 grams in the multiplied up by 100. So 50 divided by 105 times 100 exists 47 0.6% by mass of ethanol, and we could do the same thing for the water percent mass of the water is equal to the mass of water, which is 50 0.0 grams. And again, the total mass of the solution is going to be 100 and five grams times 100 and again, we're gonna get 47.6%. So for the percent mass of sugar to do the same thing and, um, 30% in my house, uh, that sugar and that is equal to the mass of sugar, which is 5.0 grams divided by the total mass of the solution, which is 105 grams multiplied by 100 and we get four point seven 6%. So now we found the percent prime ass, um, where the mass percent of each of the compounds in the mixture, but were asked how many grams of the mixture I should one take in order to have 1.5 grams of sugar. So that is asking us how many grams should be solution be if we have 1.5 grams of sugar knowing that the percent massive sugar is 4.76% so to do is we're going to use the same equation. But we're going to be solving for massive the solution instead of the mass percent massive sugar. So we know the percent mass of sugar to be 4.76% and we want 1.5 grams of sugar and we're solving for the mass of the solution. So I'm just going to write X for unknown variable, and that's multiplied by 100. So first of sulfur ex im going to divide both sides by 100. And what that will do is that will cancel out this term. And if I do the 4.762 by by 100 that gives me zero point zero or 76 which is equal to 1.5 grams over X. And if we rearrange this equation to solve for X, you would do 1.5 grams divided by 0.476 and that is equal to 30 one point five one grams. Um, weaken around this to about 32 grams. Uh, the solution or the the mixture. So that's how many grams of the mixture we would need if we wanted if we had 1.5 grams of sugar and now we are going to the same thing. For the last part of the question, we're asked how many grams of the mixture should one take to have 10 grams of ethanol? So we know the percent mass of ethanol is 47.6% and we want 10 grams can we're trying to solve for the massive mixture, which is gonna be X multiplied by 100. So again. I'm going to divide both sides by 100 so that we can isolate for X. This cancels out. Here we get 0.476 which is equal to 10 grams over X. On. If we rearrange for X, we would get, uh, 10 grams divided by zero point where 76 And that gives us about 21 grams after we round. So that's how many grams of mixture we would need if we had 10 grams of ethanol slow her answers, and these are our answers for the first part of the question.


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