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Review Conslants Penodic Tablegaseous solution of methane and propane having mass of 4.00 g was burned excess oxygen: Twenty-Iive percent = Ihe heat produced was us...

Question

Review Conslants Penodic Tablegaseous solution of methane and propane having mass of 4.00 g was burned excess oxygen: Twenty-Iive percent = Ihe heat produced was used to raise the temperature 114 g liquid water Irom 0.0 "C to its boiling point at 100.0 ~C.Par AWhat the mass methane in the mixture? (Aroub - H for methane -802.5 kJ mol and for propane 2043.9 kJ molExpress your answer with the appropriate units.MmcthanrValueUnitsSubmitPrevious Answel s Request AnsWefIncorrect; Try Again; atte

Review Conslants Penodic Table gaseous solution of methane and propane having mass of 4.00 g was burned excess oxygen: Twenty-Iive percent = Ihe heat produced was used to raise the temperature 114 g liquid water Irom 0.0 "C to its boiling point at 100.0 ~C. Par A What the mass methane in the mixture? (Aroub - H for methane -802.5 kJ mol and for propane 2043.9 kJ mol Express your answer with the appropriate units. Mmcthanr Value Units Submit Previous Answel s Request AnsWef Incorrect; Try Again; attempts remaining Part B What the mass propane in the mixture? Express your answer wlth the appropriate unlts mpropane Value Units Submit RcquesL AngWc



Answers

The combustion of methane gas, the principall constituent of natural gas, is represented by the equation $\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1)$
$$\Delta_{\mathrm{r}} H^{\circ}=-890.3 \mathrm{kJ} \mathrm{mol}^{-1}$$ (a) What mass of methane, in kilograms, must be burned to liberate $2.80 \times 10^{7} \mathrm{kJ}$ of heat? (b) What quantity of heat, in kilojoules, is liberated in the complete combustion of $1.65 \times 10^{4} \mathrm{L}$ of $\mathrm{CH}_{4}(\mathrm{g})$ measured at $18.6^{\circ} \mathrm{C}$ and $768 \mathrm{mmHg} ?$ (c) If the quantity of heat calculated in part (b) could be transferred with $100 \%$ efficiency to water, what volume of water, in liters, could be heated from 8.8 to $60.0^{\circ} \mathrm{C}$ as a result?

Mhm. Yeah. Okay, So this is a pretty large question, basically, it states few propane right, will combust. So the first question is too right down the balanced equation for the complete combustion of the property death, death. So, let's start with that. So, normally, right, we have the propane gas as a gas face and the combustion will be involving oxygen, right? Which is also gas face. And for the complete combustion, the product the usually carbon dioxide as gas face and water as liquid face. Now, we can try to balance the equivalent you relieve for combustion reaction. The easiest way to balance direction is to look at the fuel itself, right? The combusted reaction itself first. So let's look at the field, right? The propane. So in the propane molecule there's three carbons. Right? So for the product, we will say, right, there would be three carbon dioxide And in the propane there are eight hydrogen. So for the water as a product, right? Because each water has two hydrogen, so there should be four waters as product. And then we can try to calculate how much or how many oxygen right in the product, which is for carbon dioxide. There are three times two. And for the hydrogen for the water, right? There are four times one, so 10 In total oxygen in vote interaction. Right? So for the reactant there should be five oxygen molecules because each oxygen molecules has true oxygen. And this is a balanced equipment. Now we can look at the second Ukrainian question which is calculated the modeling of air at 25°C.. And one atmosphere that is needed to completely Combust 25 g of propane. Yeah, yeah. So this question actually related to the story symmetry of the chemical reactions. So it says it will combust 25g per pain based on that. We can actually calculate the more of the propane right, consumed in the combustion, which is the mass of the propane over the more mass of the propane And the masses 25g. It's given in the question and the more mass we can do the calculation right? There are three carbons. So it's three times 12 pause a hydrogen. So it's eight times one. It's grams or more. And the calculated results is around Yeah. Mhm. Let me say it's actually around. Yeah. A point fi 68 more. Right. And according to the equation, right? Every time you consume more and more simple pain, you will consume five more oxygen. So based on that, we can get relations right for the reaction the most of the oxygen in direction, we equal five times the most of the propane in the reaction, Which is five times .568 More. Right? And it's equal to two points. It's for one more. And based on that, we can now calculate the mass of the oxygen that needed for the reaction, which equals to the more mass of the oxygen times the moors. That was oxygen. The Morris of the map more more massive. The oxygen is certainly true. Right, grabs for more. The morse of the oxygen is 2.8 for one more. And this is actually equals two 90 0.9 g. So the volume of the oxygen. Then we'll be right. The mass of the oxygen over the density of the oxygen gas. So, plugging the number, The mass of the oxygen is 90.98g. And the density of the auction is also given the says right one liter. So the oxygen gas contains 0.275 um grams of the oxygen. So it's actually 0.275 g per liter. And this would be equal to Mhm around 330.3 Features. And this is the answer. So now let's look at the question. See So the questions say, is um questions that gives the combustion gives the uh entropy of the combustion of the propane and also keep the formation energy of the um the water and carbon dioxide. And it asks you to calculate the heat of formation with the propane. So, let's write down all the information we have. Right? So, first, it gives right the combustion entropy entropy. With the combustion it was the propane. It's equal to -2219.2 killed your for more. And we can first we need to write down its corresponding reactions right? Which we actually already ah Roldan it's here. Right? So that's just write it down again. Right? We have propane combust with oxygen and forms carbon dioxide and water. And the second piece of information is the formation energy of the water. It's equal to Miners. point a key to joe for more. And we can write down into corresponding reactions right? Which? Sorry? Which is hydrogen As a gas face react or his oxygen. And we're forms the water. And the 3rd pieces of the information is the formation energy. I was a carbon dioxide equal to -393.5 kHz Joe for more. And the corresponding reactions, Okay, is this right? The combustion of the carbon into carbon dioxide? And these are three pieces information we get and what we want to calculate, right is the formation energy of the propane. This is what we want to tackle it and saying we need to write down its corresponding um reactions. So since it's a formation energy, right? Or entropy of the formation, So the product should be the propane and reactant should be the pure substance which is carbon right? And hydrogen. And we need to make it balanced three carbons, right? And eight hydrogen. So it's four ha jin molecules. So through the observation. So you queer in four, right? The camcorder queer and four actually equals two. You can say right, He goes to Chemical Aquarium four equals 2. There are four hydrogen. You say there are one hydrogen so equals two. four times the chemical equation too. Plus three times You can call your question. three miners a chemical aquarian war and same right for the and therapy. Then the S. O. P. I. Was uh Propane will be equal to four times good entropy information entropy of the hydrogen plus three times the entropy of formation of the carbon dioxide. The miners. The entropy of the combustion. That was the answer. Uh the propane. And we can plug in the numbers here, which is four times minus 285 point a kilo job more. Right. Plus three times miners. 390 3.5. So here I will not write down the unit but usually it's necessary. And miners Right? This combustion energy which is -2200 and 19 point two. And you will find the calculated results will be around Um miners 100 and 4.5 killer joe for more. And this is the answer. So now we can move to the question t Assuming that all the heat released in burning 25g of propane is transferred to four kg of water, calculate the increased the temperature of the water. So this is a a confirmatory problem. Right? So, first, what we need to do is to calculate right the heat transferred through combustion of the propane, which will be equal to right the morse the propane used times its combustion entropy of the propane. So it says by 25 g of the propane will be combusted and the morse of the propane will be used is actually calculated in the question be which is here? Right .568. So we can write down here The more so the propane is .568 and times Yeah, the entropy of the combustion of propane which is already given right. It's give me the question which is um miners 2,219 2 And this equals two -1.26 times 10-3 killed. You remember it's killed um material. And it's a color geometry problem. So we're no right. The heat change for the solution will be minus sign he change of the reaction. So this will be the same. And then we can also write down the expression of the key change of the solution, which is right uh specific heat of the water times the mass of the water and times the temperature change. And the specific heat of the water, which is 4.184 clams as no grams. But joe's right, joe program particularly. And the mass here he says, used like four kg of water. So it's 4000 grams of the water. Right? And time scale that he which we don't know. It's what we want calculate right? And this We equal to this right? We equal to 1.26 times 10-6 Joe. Now it becomes joe, right? And we can use the calculator to do the calculation will get. Temperature change is equal to 75 .4° C. And this is an answer of the question.

Hello there. And welcome to problem number 85. Um, this problem is gonna be a little bit long because it does have four parts to it A, B, C and D s. So just bear with me and we will work through all four parts together. Started out for part A. They tell us we are starting with propane, and it is going through a combustion. It is combusting, and they would like us to write a balanced equation. So propane is C three h eight. Combustion means it is reacting with oxygen. He had the two products of a complete combustion are always going to be carbon dioxide and water. Okay, To balance this, I am going to need three carbons on the product side since I started with three. Hey, and I'm going to need eight total hydrogen. That gives me six plus four, which is 10 oxygen on the product side. So coming back over to the react inside, I'm going to add a five. And this is our balanced equation for the combustion. Okay, in part B, they want to know how many leaders or what the volume of air is that would be needed if we're trying to react 25 g of the propane and they tell us what percentage of air is oxygen, they give us a conversion there. So this is going to be a conversion problem. I'm gonna have to do some strike geometry, starting with my 25 g of C three h eight. I'm gonna go ahead and convert first two moles of C three h eight that I will use the mole ratio to give me two moles of oxygen. Um, Then I will use theme Mueller Mass of oxygen to figure out how many grams of oxygen are needed. And then I'll use the conversion that they gave me. That tells me how many grams of oxygen there are per leader of air. So that's my plan. Let's go ahead. But that into motion first thing I'm doing is converting from grams of propane, two moles of propane. I get a molar mass of 44.9 g of propane in every mole of propane. When I look at the periodic table and and the molar masses together for C three h eight, then using the mole ratio from the balanced equation, I see there's a coefficient of one in front of the C three h eight. I am interested in finding out information about the other reactant the oxygen and there's a five in the balanced equation in front of that. So that means for every one mole of C three a. J, I will need one mole of oxygen. Now I want to convert those moles of oxygen 2 g. So in every mole of oxygen, since it is 02 I need two times the molar mass of an oxygen atom. So two times 16 gives me 32 g of 02 And then finally, in the problem, they told me that zit there are 0.275 grams of oxygen in every leader of air. So that's just a direct conversion that I picked out of the problem. Alright. Looking at what cancels here, Grams will cancel, moles will cancel, moles of oxygen, will cancel, grams of oxygen will cancel and I am left with an answer. In terms of leaders of air, we are allowed three significant figures here, which means I have 330 with that zero being significant leaders of their and that would be my answer for part B in part C. They want us to calculate the heat of formation for this reaction. Remember the heat of formation? According to Hess's law, If we take the heat of If we take the NFL peas for the products and we subtract in this case the heat of combustion for from the reactant since we will get the heat of formation. So let's go ahead and do that. So I Delta h of formation I am going to take first the products. There are three moles of CO two and in the product and the problem. They tell us that there are negative 3 93 0.5 killer jewels for every mole. All right, so that is for the carbon dioxide. The other product is water. There are four moles of water. According to the balanced equation, take four moles of the H 20 times the NFL p value that they gave us in the problem. The heat of formation is negative to 85 0.8 killer jewels Permal. All right. This'll will give me the value for the products. Now I need to subtract the react. It's they give us the heat of combustion for the propane. There's only one mole of propane. According to the balanced equation, he and the heat of combustion for that is negative. 2219.2 killer jewels from all okay, Looking at the oxygen. Since it's a gaseous element, it's heat of formation is a zero. So I do not need to add in the oxygen. So this is my equation. How it's time to do some solving, grabbing my calculator. I'm gonna do this in a couple steps. I'm going to simplify. First of all, I'm going to calculate my products. I get negative 2300 and 23.7 killer jewels for the products. And then I am subtracting the reactant, which is negative. 2219 point to kill a jewels. Um, two negatives, of course. Make that a positive sign. So finishing this out? Yeah, I get my delta heat of formation to be negative. 104.5 killer jewels for every mole of C three h eight that combusts. And that would be my answer for part C. We are almost there. One more part. Alright. For part D in part d, we are trying to determine the change in temperature if we use 4 kg of water. So how much? How much will the temperature of that water change if we combust? 25 g of propane. Okay. Yeah. All right. Let's see here we are trying to figure out change in temperature. Looks like the equation I'm going to use is the amount of heat produced. He's going to be equal to the I'm sorry, that should be a little lower case C because I want specific heat here. So the specific heat of water times its mass in grams put him times the change in temperature. That's the equation I want to use because I can look up the specific heat of water. I have a massive water given, and I can calculate you based upon the information that we have in this problem. So let me go ahead and calculate que I need to calculate. Q is given. Our Q is going to be the heat energy produced by the combustion of this methane. That is, of course, given to me already in kill the jewels per mole. So I need to figure out how many moles of propane I have so that I could figure out how Maney killer jewels will be released. So starting with my 25 g, I need to calculate moles. So do that. I'm gonna take my 44 09 g, which is my Moeller massive C three h eight in every mole. Yeah. Then I'm going to take that heat of combustion we were given for propane. We were told that it is 2200 at 19.2 killer jewels. Permal. Since I'm ultimately determining how much energy is absorbed by the water, I'm going to use the positive value of that. Of course, the negative value tells you how much is given off. Now. I'm interested in how much is absorbed. I'm sorry. That is Permal got ahead of myself a little bit there since I was talking. All right, so that is per mole of the C three h eight. Finally, what I want to do is I want to express killer jewels in terms of jewels, because I know my specific heat is always given in jewels, so I'm just gonna convert. There's one killer Jewell is equal to 1000 jewels. This is going to give me an amount of heat in jewels and I get a pretty large number since I converted it to Jules. Alright, I get that many jewels. I am ready to plug all of my values into this equation and solve four Delta T. Let's go ahead and do that Cube. That's the amount of heat released or absorbed in this case. So that's the number of jewels I just finished calculating. That's gonna be equal to the specific heat of water looking that up in the table. It is 4.184 Jules program Degrees Celsius. Notice. This is Ingram's. So I want to express my 4 kg of water in terms of grams. So 4 kg doing a quick conversion here in 1 kg, there are 1000 grams finally time still to t so I can now solve. I could go ahead and solve this now for Delta T. And I find that the change in temperature three only unit that's gonna be left because Jules and Grams are going to cancel the only only unit remaining are gonna be degrees Celsius, and it is 75.2 degrees Celsius. And that is our answer for part D. Hey, thanks so much for sticking with me through the entire problem. Oh, I hope you found this helpful.

In this question The balance in reaction, which in can be written as 4.4 c three h eight plus 4.6 C h four plus a theoretical off 02 plus 3.76 and to will give us 1.8 c 02 plus 2.8 edge toe plus issue radical times 3.76 And to by balancing the oxygen, we can get a theoretical equals 3.2 The second part. This is the first part in the question. Second part. Partial pressure off water vapor TV equals h en ough h 20 over, and total times be total equals 16.84 kilo. Askin the two pressure or do you point temperature off the product of gasses? T DP equals de saturated equals 56.2 degrees soldiers from Table E five and import to see in this question the heat transfer for this compulsion process is determined from energy balance. So where were the temperature change in temperature? Delta T equals 1 50 minus 25 equals 125 degree surges, which is equal 1 25 Calvin, then using the values given in the tables we can calculate negative que out, which would be equal negative. 1233 700 Qi Lu Xun A key newman Off the fuel we can remove This negative was just negative and for the heat, transfer it off. 140 1000 Kluge in per hour then and fused equals que dot out over Q out. So we have opened 1135 fuel. They're our and the Mueller mess off a few mixture is yes, off the fuel equals 0.4 times 44 plus 0.6 times 16 equals to 7.2 kilograms Their kids the math, florid and fuel equals and fuels I'll supply and Fuge equals 3.87 Killer crime where our in the airfield ratio f equals mass off air over math, off fuel. So we have 16.24 telegram off air and kilogram Off yours Now the mess florid off air and is a punch. See what diamonds F, which is equal 50.1 kilograms at our

The explanation of the solution is that now for the part A that one more off meeting. Liberals 890.3 Gradual off heat on conversion. Here the Mosque of Methane will be that 2.80 multiplied 10 to re power seven KJ multiplied by one more CS two divided by 890.3 Closer multiplied by 16.0 for G ch four divide by one more off ch four multiplied by one club from divided 5000 Drum and help we get the moss off is 504 Hello, Graham ch four This is the muscle Maintain now for the part B Her first We did remind the moles off ch four present using the ideal gas equation We know the ideal execution p we call to an arty and here it is an equal toe. You divide by rt here putting all the values in the equations And by calculating this, we get the value off and is 696 more off ch four Here we can go to the mall off CS four. So the heat energy is equal to six 96 small ch four multiplied by 890.3 Closer divided by one more off ch four And here it comes toe minus 6.20 multiple by 10 to the power five Close you off heat energy. So now we know that the heat is equal Toe mosque multiplied by specific hate multiplied by temperature change. So here the volume of fortress mask developed by density. So this will be cool toe he tickle toe hate divided by specifically to multiply by temperature difference my brother one divided by density And here by putting all day values in the question calculating this help we get the value is 2.90 multiplied by 10 to the power six am All 2.90 multiplied by Can't you power three leader s door So this is the complete solution Off the answer in the daily step by step. Let's go through this. Thank you


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