in-the-laboratory-you-run-a-procedure-for-the-reaction-described-in-exercise-5-and-produce-29-g-of-2

Name: in the laboratory you run a procedure for the reaction described in exercise 5 and produce 29 g of 2
Uploaded: 2020-10-18
Duration: 1 min 46 s
Channel: Numerade
Description: In the laboratory, you run a procedure for the reaction described in Exercise 5 and produce 2.9 g of silver chloride. What is the percent yield for your procedure?

Question

In the laboratory, you run a procedure for the reaction described in Exercise 5 and produce 2.9 g of silver chloride. What is the percent yield for your procedure?

David Collins · Accepted Answer

after that. That very lengthy question in question. Five. The question six is just to follow up a short follow up to question five. In this case, it asks us now to determine percent yield percent. Yield, as you recall, is simply equal to the actual mass we determine in the laboratory. Over the theoretical in the theoretical is what we calculate or what we calculated in the previous problem. Times 100. So what did we calculate in the previous problem? As you recall, we calculated a mass, a theoretical mass of a GCL that could be created from 6.3 g of silver nitrate and 4.5 g of sodium chloride to be 5.31 g silver chloride. So to calculate percent yield, we&#x27;ll take the actual mass which they have given to us. We can&#x27;t calculate the actual mass. This is what we got in a laboratory. So they have to tell us we get 2.9 g. We should have gotten according to our calculations, 5.31 g. We multiply that by 100 and we get a 54.6 percent yield

Matthew Lueckheide · Answer

So this is another yield question. Right? So they&#x27;re telling us that magnesium chloride is decomposing into magnesium metal, and, um, we start with 185. Sorry, we recover 185 grams of magnesium, but we started with 100,000 grams or one kilogram, um, of magnesium chloride. Right. So you&#x27;ll notice that I didn&#x27;t finish the equation, right? Because I wanted to do it with you. Um, so when you have magnesium core of any metal core, when they decompose and you form the elemental metal is what we call this magnesium by itself, right? So the balance, the equation, you&#x27;re always gonna make Korean gas as a product. It&#x27;s not important for this, um, the math at all, But I just wanted to, uh, practice writing or predicting products. Right. So we start with 1000 grams that we give the theoretical yields so more convert when we&#x27;re calculating percent yields. Right. Um, it&#x27;s the actual amount that you collect, which in this case, is gonna be the 185 frames over the theoretical amount of front. That&#x27;s forms based on the math that we&#x27;re going to do right now, right? So 1000 grams of magnesium chloride. Right, So we&#x27;re gonna go moles of magnesium chlorate. And the way that we do that is the muscular. Wait. So, based on my calculation, um, the more here wait is 95.21 grams of magnesium chloride that&#x27;s gonna cancel out, right? It&#x27;s a 1 to 1 ratio so we can get one mole of magnesium over one mole of magnesium chloride, and then the mass of magnesium metal will get us two grams of magnesium. So that&#x27;s 24.305 If he was my desk malted Harry through this is grams of magnesium, right over one mole of magnesium. So mole is cancelled malls. It&#x27;s now we&#x27;re left with grams of, of, of magnesium. Um, and when you round that out to a few decimal places, you should get 255 0.28 grams. So when we calculate, the percent yields right, So the percent yields, um is going to be, ah, it says that we collected 185 grams, right? And how much is apparently made in the reaction based on our math is 255.28 grams. Remember it. So frak is a desk. Sorry. We&#x27;re going to get a decimal, so we have to multiply by 100 to get the percent. And when you do that, Matthew should get 72 0.5% yields.

Joshua Speer · Answer

So the theoretical yield is the yield that you would get if your women in region that&#x27;s the regent. You have the least amount of reacts completely and only forms your products and no side products or anything like that. No un reacted materials left over on DH, then the percent yield is the yielding actually get from running the experiment for the reaction divided by that theoretical yield. So if he only made half a CZ much as you thought you were going to make them, your percent yield would be fifty percent.

William Mills · Answer

this question asks us to find the percent yield reaction in which 25 grand&#x27;s were isolated. But the theoretical yield was 81 grams percent yield healed, he calls actual. That is a sin ti over theoretical. I repeat, that is th e. And in this equation the student got 25 grams of actual ute and the expected yield was 81 grams. You do this, you will find that the answer is 31% said the percent yield for this reaction of the student did was 31%.

Kim Trang Nguyen · Answer

Okay, Father Problem. 1 41st we have toe Nick this month. That&#x27;s money. Why do something Stop problem by this disease. What a way we can! So so last sorting where I had a mom and bought a single or I get be more. So you let the Russian off the mystery with a set super nitrate when they say such of when I tried that mean sodium fluoride Both asking a lot. I convinced me to actually soup when I drink and no more. And the after we die in question, we have to find a more radio with more right or money problems in the many rivers rivers problem. So now I do. I think you&#x27;re familiar with this. So what? Most of the ball, right? I got a more civil 90 s and that a more civil All right recipe day The same for me, Mom. Potassium chloride. We have be mom. What? That&#x27;s him precipitate. And now we know and we more time on a must. That&#x27;s why I&#x27;m off civil colorize equal Moloch mas of civil. All right, I&#x27;m time Optimal arrive possible will rise up in the question A and be the first question is sick ambition a less theme? All right. And that we have on We spend this from this form, you know? So we get one for three for a one off one for three by 58 plus one. Part 35.5. Big equal. 1.913 grams of ala. Right, This number. Davy was quick in the problem. And lest the similar way with company for the mass of mutual, but doesn&#x27;t learn I and Castle. So the ball, right. So must also what I saw the mother mother for the right time for them also right and mass support. That&#x27;s a lotta equal model possible. That&#x27;s all right. Come. Far more observant. So what? That&#x27;s a blow, right? So from this question, we have 58 boy eight and time for 74 foot. So the 4.5 b, will you and be so we&#x27;re with another units that easy, though That&#x27;s much out how I shall most in the fire. And now I has set off a question. I wanted to do one right here, one in possible right, and be too far the muscle. Mitchell, I have this question so I think you&#x27;re familiar with the system. And you could your calculator. You can buy a nd after you get a a That mean that my officers right on your tie for the mother mouse About Mariah. You that a boy? Three a one round. And I said You come with a for the mass offs possible right is that they don&#x27;t need to pound a month for that&#x27;s not drive us. I call that here too, because the only i percent so after that you got percent off sudden chloride divided by the mass of the picture. You left 23 poison for 75%. And because they only tools to something so one get was already you can get and I want to present minor another percent u had a Baz enviable. That&#x27;s simple. Alright. And here I have I you see, maybe our and so I needed different from gonna be on about formed about as off another professor. So here, because I use it for housing for people Wanna wait Mooring equals 5.533 But at some point tonight, that easy for auto populates good

Carlene Jimenez · Answer

So this question gives us this reaction between lead nitrate and aluminum chloride and tells us that eight grams of lead nitrate is reacting with 2.67 grams of aluminum chloride in this reaction is giving an actual dealed of 5.5 five grams of lead chloride, and it wants us. Its chairman was the limiting re agent. And what is the percent yield? So first, we need to determine, um, which is limiting re agent. And we can&#x27;t do this unless we have the molecular weights for the compounds of interest. Um, and that that is lead nitrate and R. Lewin chlorides weaken. Um, calculate that, um, quickly by summing the rates of each adam which make up the compound. And we&#x27;ll find that the molecular weight of lead nitrate is 331.22 grams per mole, and the weight of lead chloride is 133.33 grams per mole, and because we&#x27;ll need it later, will also just calculate the weight of lead chloride. And that is 278.1 grams per mole. Um, so now we can determine what is limiting. Um, So first, let&#x27;s see how many mole of lead nitrate is in eight grams of it. And we can do this by referencing our conversion triangle, which gives us the relationship between the mass of compound and grams, the number of moles within that, um, particular weight of compound and the molecular weight of that compound. And his conversion triangle tells us that to find the number of moles, weaken, divide the weight by the molecular weight. Um, and we can do that for lead nitrate and divide its weight in grams by the molecular weight of lead my trade, and we&#x27;ll find that eight grams of lead nitrate contains 0.2 for two mole of blood nitrate. And we can do the same for, um, aluminum chloride. Um, so we will divide the weight of aluminum chloride Ingram&#x27;s by the molecular weight of aluminum chloride and grams per mole. Don&#x27;t we&#x27;ll find the 2.67 grams of aluminum chloride contains 0.2 mole of aluminum chloride. Um, but this doesn&#x27;t really tell us which is limiting. What we have to do is compare um, how many of one compound is react is necessary to completely react with the other. So let&#x27;s just choose, um, from, um let&#x27;s just choose from aluminum chloride. Um and we want to know how many mole of lead nitrate, how many wool of lead nitrate is required to react completely with our 0.2 mole of aluminum chloride. And we can do this by multiplying our number of mole of aluminum chloride by the conversion, um, we can get that ratio from our balanced chemical equation. Um, we know that for every two mol of aluminum chloride consumed in the reaction and we can tell this just from the coefficients for every two moles of aluminum chloride consumed in the reaction, three mol of lead nitrate is also being consumed. Um, so the number of mole of lead nitrate we need to completely react with 0.2 moles of aluminum chloride is 0.3 mol of Lead Ni tricked. Um, and if we start to compare the values that we have here, we need 0.3 mol of lead nitrate to completely react with our given bowl of aluminum chloride. But we only have 0.2 for two mol of lead nature. So this value that we have actually for lead nitrate is less so. Therefore lead nitrate is limiting and our reaction will only go as far as to consume all of lead nitrate as the aluminum chloride is in excess. Now, this question also wants this determine what is the theoretical yield of lead chloride, given that the actual yield of lead chloride was 5.55 grams and they also want us calculate the percent yield. So now that we know what our limiting re agent is, we can determine what the theoretical yield is. Um, since you know that all of zero point 0 to 42 moles of lead nitrates being consumed this reaction, you can begin determine how many mole of lead chloride will be made from the amount of mole of lead nitrate. Okay, so we know that for every three mol of lead nitrate made sorry for every three mol of lead nitrate consumed in this reaction, we know from our balanced chemical equation that three more of lead chloride is produced, um, so they are consumed and produced in a 1 to 1 ratio. So the number of mole of lead chloride produced from 0.242 MOL of lead nitrate is 0.2 for two bulls of lead chloride. But the question wants us to calculate the theoretical yield, Um, which we can do with grams. So we just have to convert this number of mole of lead chloride for moles to grams. And we can do that again by referencing our conversion triangle, which tells us that mass can be calculated from the product of the number of roles and molecular weight. So we really did. It multiplies their 0.0 to 42 moles of lead chloride with, um with its molecular weight, which we found previously to be 2278 0.1 grams. Um, grams formal, actually, um so with this will determine that the theoretical yield of lead chloride is 6.73 Grimm&#x27;s um, the question wants us to determine the percent yield, and we can do this by taking our actual yield and dividing that by the theoretical yield. And we&#x27;ll applying that value by 100 and we know our actual yield was 5.5 grams. Sorry, 5.55 grams in our theoretical yield was 6.73 grams. So we will multiply that value my 100 we&#x27;ll find that the percent yield of this reaction was 82.4 seven percent.

Problem 6 Easy Difficulty

In the laboratory, you run a procedure for the reaction described in Exercise 5 and produce 2.9 g of silver chloride. What is the percent yield for your procedure?

Answer

54.7$\%$

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