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Fe(CrOz)2 +8 KzCO3 7 02 _ > 2 Fe203 8 KzCrO4 + 8 CO2 Use the following conversions: 1 mol MW in g (2 dp) = 22.414 L gas at STP What is the theoretical yield if ...

Question

Fe(CrOz)2 +8 KzCO3 7 02 _ > 2 Fe203 8 KzCrO4 + 8 CO2 Use the following conversions: 1 mol MW in g (2 dp) = 22.414 L gas at STP What is the theoretical yield if 2.90 g of Fe2O3 (159.70 g Fe203) are formed from reaction of 10.0 9 of Fe(CrO2)2 (MW -223.85 g Fe(Cr0212)?Theoretical yield

Fe(CrOz)2 +8 KzCO3 7 02 _ > 2 Fe203 8 KzCrO4 + 8 CO2 Use the following conversions: 1 mol MW in g (2 dp) = 22.414 L gas at STP What is the theoretical yield if 2.90 g of Fe2O3 (159.70 g Fe203) are formed from reaction of 10.0 9 of Fe(CrO2)2 (MW -223.85 g Fe(Cr0212)? Theoretical yield



Answers

What is the percent yield if 108 mg SO2 is isolated from the combustion of 82.7 mg of carbon disul“de according to the reaction: CS2 3O2 ¡ CO2 2SO2?

To find the theoretical yield of carbon dioxide. Let's right the chemical reaction to which they are referring, and there's no need for coefficients. It is balanced at, as is well, then take the 10 grams of carbon that they tell us that is reacting and converted into molds. Carbon and then moles. Carbon levels, carbon dioxide recognizing it's 1 to 1 and then moles carbon dioxide, two grams carbon dioxide and we get 36.6 grams carbon dioxide as theoretical yield carbon dioxide.

This problem is also a limiting reactant problem where we are given to amounts for two different reactions and we need to figure out which one is the limiting reactant and then from the limiting reactant to determine the theoretical yield and ultimately the percent yield. The balanced chemical reaction is two moles of iron. Three oxide reacts with three moles of carbon to produce four moles iron And three most carbon dioxide. So if one of the reactant is iron, three oxide and we have 12.5 g of it, we can convert those grams into molds by divided by the molar mass of iron three oxide than knowing the moles of Iron three oxide required to produce four moles of iron. This being the ST geometry, we can calculate the molds of iron produced, then multiply by the molar mass to convert the moles iron into grams iron and we get 8.74 g iron produced from 12.5 g iron three oxide. Next, we'll do the same thing with carbon. We have 2.6 g of carbon, which we can convert to moles carbon by dividing by the molar mass carbon. Then, when we know the moles carbon, we can calculate the moles of iron With the 3- four strike geometric relationship. Then when we have moles, iron will multiply by the molar mass iron to get graham's iron And if all 2, 6 g of carbon were consumed would get 16 g of iron. Therefore, iron three oxide produces the least amount of iron. This being the theoretical yield 8.74 g and iron three oxide producing the least amount of iron is the limiting reactant Percent yield will be what we actually get, which is provided in the problem, divided by the theoretical multiplied by 100 and we get 80.1 yield.

Hi, everyone. So first world will date given data No volume officer. True, they're given on there is equal to to 85 point fireman. So this urge to convert into leader by dividing by 1000 So bye in 285 5 liter then one Lima for true they provided, of course, This is a barge volume off a true 1. 58.9 a mill that is equal to point one for you. 89 liter. What is the limiting reactant they ask limiting reactant. Okay. Thus secondly, dust hieratic leadoff Eso three Their article it off a so three Hey, something. And in p parts. They provided us with a limb off. Assad went up there. So three, That is two point 810 for you, Graham soap. So first to parade reaction. Oh, you know toys. A sort of gas less or to gas. Gas. That is so three guys. No, fast. You have to calculate. In the little mall suppressant, there is equal toe TV a found charity This from General the execution p visit contain ality. And that's why again is equal to TV a polarity. Uh you know, they mention it spp so pressure at https one even. We'll mention this in the given data. That is pressure at http is one atmosphere. Find temperature is 2000 and three find 15. Calvin. Okay, so this is one into Varlamov. I saw the point to hate 55 do I do? By all zero point 0 82 in tow. Temperature to 73.15. So that is equal toe point 2855 divided by 22 wine. 39 83 83 therefore Yeah, enough. I saw too. That is equal to 0.0 12 765 balls. So the soul molds office or two. Then second step is to calculate mall. So for two enough for me And there is a call dot PV upon our tree There is a cultural pressure is one then Balinese point run fire 89 divided by art 0.0 82 in 2 to 73.15. So that is equal. True points 1589 Defended by 22 point 39 83. Therefore And now for truth, that is equal to 0.7 094 moles now charge on. Let us compare the related quantities off the creation in terms off mold comparison. Enough related on today's computing's okay, off creations, you know, zero point 0127 65 this present for to most office or three so divided by can see True, it is smaller than, uh, 0.709 form also oxygen onda four This comes 0.0 6385 heard five more Officer Toe Moeller off assault do which is having are low concentration bank 0.7094 more lawful to more law for two and this in funds night. Since oxygen is more, more on de Soto is less I thought to is less Therefore I saw two is the limiting the agent is that limiting the agent between creation? Yeah, no fault from equation We're seeing number of moles off eso three a number of moles Office Soto They're equal number of moles Office Soto It does V zero point 01276 months and therefore and what theoretical models office or two hair article Also Pessotto also has a three. Uh, that is 0.1276 malls Then we'll calculate theoretical in office. So three fit, therefore tear article. He'll love assault you eso three is equal to a number of small soft s o t him Toe Moeller Months off assault So member of Molson pesos 0.1276 into model Massa s a 3 80 point, not six now, four tear article, field office or three sure is equal to one point 0 to 1 grand. One point yeah, 0 to 1 g nos. Now be the ask a beat bark. They ask percentage, you can say actually in Love has sold three actually in off a sultry they provided us as 2.805 a gram at Estee Pee had a steep e soul percentage inch off a saw three That is equal truth. Okay, actually, upon theoretical in is into, um danged So that is equal toe actually lays 2.80 fine upon theoretical. India is one point 0 to 1 in 200 therefore percentage yield off. I saw three is calculate rash to 74 points, 7%. Then be a part If if 0.80 off for you. Graham is the trapped eso three trapped and so three at sleepy and therefore percentage each awfully action. That is equal True 0.805 g defended by three articles in 200 and therefore percentage is off. Reaction is equal to 78.84%. 84%. Thank you.

Okay, so we this is a percent yield problem. And ah, we are given the combustion of octane once again, and we're told that we observe the formation of 1.90 times 10 to the third grams of co two. Um, we have to figure out the percent yields of this reaction. And how many grams, um, are possible to be formed, Right? So that's that's the equivalent to the, um, the theoretical yield of this reaction. So we're gonna start with one leader of octane and we're gonna have to get to grams so we can get two moles, right? So the density of octane is 20.700 grams per millimeter. So let's first convert believers, the middle leaders using the prefix from an earlier chapter. Right? So if every 1000 milliliters is one leader or, um, another way to express that is for everyone. Mill leader. There's 10 to the minus third leaders. It's the same thing. According to that table in the first chapter, I think it's the first chapter of this textbook. Um, right now we're no leaders. Now we're going to use the density to get to Graham's ransom 0.7 zeros your grams for every one millimeter. Then we could get two moles. So I've calculated the weight of octane to be 114 0.23 grams per mole. So we're doing great canceling out right now. We're in bowls of octane. Very good. The leaders cancel out leaders, cancel out. Awesome. Um, so now we're going to use the conversion between the the coefficients of the balanced equation. Right? So if you balance his equation for you'll know that they're 80 two's to everyone. Uh, mole of octane. So we're gonna do eight moles of CO two for every one bowl of octane, and then grams of co two, and then we're done with this part of the problem. Right? So, uh, the molecular weight of CO two was 44.1 grams for everyone. Mole bulls Cancel out. Now we're in grams of co two. So this is the answer. The first part of the problem, which is how many grams can possibly form, right? So without rounding right away, um, doing this math, I was able to get, um for two thousands. Yeah, 21. Sorry. 2157 points. 54 grams. Right? Um, And when you ground this 23 significant figures which is needed for the answer, uh, you can report this is 2160 grams of co two. So here's my issue with these parted problems is that do you use the, um, the answer. So we have to report each answer to each part with the right number of significant figures that looks like that's three based on the numbers that are given. So we're gonna report our answer that the way you would report this on a homer problem if you have to account for significant figures is using this right. But the philosophy of answering these questions is that you take the answer before you don't want around that any point in the problem. Um, before you calculate before the last calculation and get your answer. So you have all of the you have the most accurate answer before you around, right? So if we really used 2160 as our theoretical yields, as opposed to 2157.54 um, your answer might be different, right? So I've done this. I've done both, and the answer is actually only off by 0.1% which could be big. Depending on your teacher. That could be all you need to lose credit. Right. So let me start a second notebook here, Right? So percent yield. Just to remind you, um, is the actual I'm writing like a monster theat actual products that you observe over the theoretical product that's possible to form based on the street geometry of the reaction. Right. So we were told that 19 Sorry at one point. Sorry. 1.9 times. 1.0 times. 10 to the third grams of co. Two were observed. That's the actual so aptly observed formation. And we calculated, um, 2157.54 grams of CO two. So we're calculating the percent year. We've gotta multiply this by 100. So the answer that I got was when you do when you use 21.57 you get 80 eight 0.1% yields. If you use, um, 21 sixties if you use your rounded answer from part, eh? Um, you you you you would get 88.0% yield. Um, So I, um I would suggest you can round your answer for reporting the answer, but then use the more accurate answer when calculating your yield is my opinion. So I would report the answer is 88.1. Um, if I were grading this, so yeah, that's the answer to this problem. Just, you know, significant figures are important, but make sure that you're reporting the right answer, right?


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