write-balanced-chemical-equations-for-the-combustion-of-the-fuels-in-the-heats-of-combustion-table-2

Name: write balanced chemical equations for the combustion of the fuels in the heats of combustion table 2
Uploaded: 2020-10-18
Duration: 2 min 25 s
Channel: Numerade
Description: Write balanced chemical equations for the combustion of the fuels in the heats of combustion table on page 525. What generalization can you make about the amount of oxygen required for combustion of hydrocarbons?

Question

Write balanced chemical equations for the combustion of the fuels in the heats of combustion table on page 525. What generalization can you make about the amount of oxygen required for combustion of hydrocarbons?

David Collins · Accepted Answer

this question is actually really long. Question. End up having toe right and balanced several combustion reactions for the fuels that air found on Page 5 25. So let&#x27;s get started. This is mostly just a balancing exercise. Hydrogen reacting with oxygen to produce water. We&#x27;ll have to hydrogen one oxygen and two waters. Carbon reacting with oxygen to produce Carbon dioxide just has one of everything. Methane, where it needs to react with two oxygen&#x27;s to produce one carbon dioxide and two waters. Ethane needs to react with oxygen, two of them seven oxygen&#x27;s to produce four carbon dioxides and six waters. Butane will react with oxygen to produce carbon dioxide and water. We&#x27;ll get two of them reacting with 12 oxygen&#x27;s to produce eight carbon dioxides and 10 waters. And we&#x27;ve got heck saying reacting with water, two of them reacting with 19 waters to produce 12 carbon dioxides and 14 waters. Then we&#x27;ve got octane and we begin to see a pattern. We see that we&#x27;re increasing the number of waters by four every time. We&#x27;re increasing the number of carbon dioxides by four every time, and except for this one right here, we&#x27;re increasing the number of oxygen&#x27;s by six each time, while we keep to coefficient on the hydrocarbon. And for methanol, we get one or two methanol XYZ reacting with three oxygen&#x27;s to produce to carbon dioxides and four waters ethanol when ethanol reacting with three oxygen&#x27;s to produce to carbon dioxides and three waters. Now we&#x27;ve got butin. All one butin all reacts with six oxygen&#x27;s to produce four carbon dioxides and five waters and then glucose one glucose reacts with six oxygen&#x27;s to produce six carbon dioxides and six waters.

Reed Mckee · Answer

all right. Today we&#x27;re gonna be doing some combustion reactions with hydrocarbons were given these four hydrocarbons on the left, and we&#x27;re gonna need to write a balance combustion equation for all of them. So first, here we have a thing which could, see, has two carbons and four hydrogen. So for the first one, we&#x27;re gonna rock E to H four. Gonna be a gas. All right, so we have two carbons in the left, so we&#x27;re gonna need to on the rock. So let&#x27;s go ahead and put to right there. Balance that out. And now we need four hydrogen with right, because we have four on the left. Have this h two from H 20 So if we add a to coefficient right there to see that we have four hydrogen, the rot as well. Go ahead and add up the oxygen to see if we can balance it with this. 02 molecule. We&#x27;ll have 234 plus thes two, which is six. So to make this six, we&#x27;re gonna add three. And this will be a balanced equation for the first. Okay, next we have protein. So this will be the three and 123456 hot. The C three h six. Oops. That&#x27;s gonna be a gas. Okay, so way have three carbon on the left. So let&#x27;s just go ahead and put three car. But on the rock, we have six hydrogen, and we&#x27;re gonna need to make this 16 hydrogen. And now let&#x27;s see if we can balance out the oxygen. We have two oxygen, plus these six plus these three. So you see here, that&#x27;s an odd number. And with this 02 molecule, we&#x27;re not gonna be able to balance this one out, so we&#x27;re gonna have to add something else. So let&#x27;s go ahead and change. This Do can&#x27;t change the other ones. Let&#x27;s see, we have six carbon around here, so go ahead and put six carbon on this side. Now, we have 12 hydrogen now, so we&#x27;re gonna need to make this 1 12 so we&#x27;ll add a six is Well, all right. So now we have 12 oxygen, plus the six, which is 18. We&#x27;re gonna need to make this age 18 so? Well, I had a non right there, and you can see this equation is gonna be balanced all right. For the next one, we have iso butane, which is just the next hydrocarbon in them in the sequence here. So that will be C four h 10 I believe word. Yeah, it would be a gas running over. You are us on. All right, so we have four carbons on the left. So didn&#x27;t do the same thing, but four on the right. We have 10. Hydrogen will make this 1 10. And then let&#x27;s add up all the oxygen. It&#x27;s eight. Plus, these five 13. So we&#x27;ve done the same thing that we did up here. We&#x27;re getting an odd number of oxygen atoms. No, we&#x27;re gonna go ahead and change that up a bit. And what another two? Two times four. That&#x27;s eight, but eight there. 20 hydrogen. So you get 20. We&#x27;re gonna have to add a tin right there. And so we have kin, plus the 16. That&#x27;s 26 oxygen on the right side. So what? Can we multiply by two to get 26 13? Because you can see for these kind of problems. You just have to work with the numbers until you get it right. Finally with cyclo butane we&#x27;re gonna have four carbons, but we&#x27;re gonna have a different number of hydrogen, so Oh, no. Back a little too far. It was 13. Run. All right, will do e four each eight. But this one. It&#x27;s similar to the amount of carbons in the last one, but we have a different number of hydrogen since it&#x27;s in a It&#x27;s an enclosed circle right here. All right, so we have four carbons on the left side, so we&#x27;ll go ahead and do the same thing and put four on the rot. We have eight hydrogen, so we&#x27;ll make this 18 four times two. Now we&#x27;ll see if we can balance out the oxygen were four times to That&#x27;s eight plus thes four. That&#x27;s 12. So all we need is a six here. And so it he&#x27;s had the same amount of carbon that we didn&#x27;t have to change the coefficient in front of the hydrocarbon. If you see here, we&#x27;ve balanced all four chemical equations

Bin Chen · Answer

Okay, So for this question, they want us to complete the compassion reaction and then balance it for the following hydrocarbon. So what I&#x27;ve done is I&#x27;ve gone ahead. Encountered the number of carbon that Hodgins in each one of the hydrocarbons, and now we&#x27;re just gonna convert them until combustion reaction. So for plantain, we have C five age, 12 plus o two yield CEO too, plus H +20 All right, so let&#x27;s balance the carbons First. We have five carbons on the left side. We have one carbon on the right side. So in order balance that we add a five here. This gives us five carpets on both sides. Sarah&#x27;s goes Arjun. So we have 1200 on the left, and we have two hydrogen on the right, So in order to balance it out, we need a six year. This gives us 1200 words on both sides. It sounds to option. So we have one times by six to have six plus two times by five. So that&#x27;s tens. We have 16. Oxygen&#x27;s on the right in another to make the number of auction on the right 16. We need to add it. It&#x27;s here. This gives us 16 on both sides. That will be our complete balance Reaction now for Part B. We have metal benzene. So let&#x27;s right the combustion reaction for methyl benzene. So we have si seven h eight plus O to yield Co two plus H 20 So let&#x27;s start with the carbons again. So we have seven carpets on the left, one carpet on the right, and they were to make it seven on the right. We need to add a seven foot in co two just gives us seven on both sides and for hydrogen. We have eight on the left when we have two on the right. So in order to make it eight on arrival, need at a four for the H 20 so we have to time to buy. Four gives us 800 hours on both sides, so let&#x27;s bounce out the options. We have one times about forms. We have four plus two times by seventh. That&#x27;s 14 equals 18. Oxygen&#x27;s on the right. Now, in order to make the number of oxygen on the left 18 we need to add a nine in front of the 02 Syrian nine times way two times by nine gets gives you 18 and that is the complete balance Reaction now for the next part of cycle. Heche saying We have C six h 12 plus O two yelled co two plus h 20 my solicitor departments against. We have six carpets and I love we have one carpet on the right In order to make it six. We had a six in front of the CO two. But sit down the carbons. That balance effort Expert, we have hydrogen. We have $1200 left. We have true Hodgins on the right from the show. In order to make a 12 we just need to add a six and friends H 20 and then we have two times by six. Just gives us 12 virgins. So now hydrogen is balance. Lastly, we goes oxygen. We have one option times by six. This gives us six plus two times by 6 12 which equals 18 oxygen on the right in order to make the number of oxygen left 18. We just put it nine. If it off Theo too. And there the reaction is bounce. Therefore party for I saw acting. We have C eight each, 18 plus O two yelled co two plus it&#x27;s too low. Some four carbons. We have eight on the left carbons. We have one on the rights nor to make eight. We put eight in front of the seal to this gives us eight carbons on both sides. Now for hydrogen. We have 18 on the left and we have two on the right in order to make it 80 which is needed that a nine in front of the issue. Oh, which is is 18 on both sides. Now for oxygen, we have to times about 8 16 plus one times in my nine. It&#x27;s 9 25 oxygen on the right and the number of oxygen on the left is to So since we haven&#x27;t even number of auction on the left and no matter what number you multiply by, two always give you even number. We have to somehow make this an even number. So what we can do is we multiply everything in the chemical equation by two rice and nine times of my to we&#x27;ll give you 20 no. We&#x27;ll give you 18 eight times by two. Will give you 16. One sounds about you will give you two and one times by two. Well, kid, you to for this thing I was count the number of carbons. We have a tense by two. We have 16 carbons of one times by succeeding gives you 16 carbons. I have 18 times by two. Just gives you 36 carbons. Now we have two times by 18 which also gives you 36 carbons. Now for oxygen, we have one oxygen times about 18 which is 18 plus two times by 16 which is 32. Right, So 30 to plus 18 gives you 50 oxygen. Now, in order to make a 50 on the left side, we need to add more o twos. So what I do is I do 50 minus the number that we have on the left side. Right now it&#x27;s two times two. Which is for this gives you 46. We do 46 divided by two gives you 23. You had 23 to 2 gives you 25 that is the complete balance reaction. So I will. So it&#x27;s to see eight h 18 plus 25 +02 yields 16 c +02 Plus It seemed h 20

Charotte Mcadam · Answer

here. We&#x27;re just looking at the complete combustion off several Al cane chains. So are examples hit. We have a vaccine cyclo vaccine unto me Fail plantain, where we take the molecular formula off each respective molecule. And then in the presence off oxygen, we are able to come pleat lee, convert this into carbon dioxide and water. Now, the number of carbon dioxide and water molecules depends on the molecule that you are working with at the stop. And then remember that your equations do in fact need to be balanced. And so you will need to adjust the number of oxygen molecules while as carbon dioxide and water molecules are present in your reaction given the compound. But you start with in the combustion reaction.

Angelina Chavez · Answer

in this problem were asked to write a balanced chemical equation for the complete combustion of various compounds in the presence of excess oxygen in combustion reactions. We know that our equation is going to take the form of some hydrocarbon reacting with oxygen to form water and carbon dioxide. Our first reaction occurs with beauty ain, which is see for each 10 and it is in the gaseous state. We&#x27;re reacting that with hydrogen gas and we&#x27;re getting liquid water and gaseous carbon by outside. In these reactions, we almost always want to start by adding a coefficient of two out front of our hydrocarbon here so we can add our co efficient too. And we can think about how we can then balance either our carbon or hydrogen based off of this coefficient that we&#x27;ve added. So we can see that by adding this coefficient. Here we have eight carbon on the left hand side so we can add a coefficient of eat tour carbon dioxide over here. So now they&#x27;re Kurban is balanced. We can reevaluate. How we can remedy are in violence of either hydrogen or oxygen. We can fix our imbalance of our hydrogen by any a coefficient of 10 in front of our water. And from there we see that we have 26 oxygen on the right hand side so we can think about what coefficient can be added here to this oxygen gas. To give ourselves an equal number of oxygen on both sides of this equation. To achieve our desired number of 26 oxygen, we can add a coefficient of 13 out in front of our oxygen gas. In this next reaction, we are dealing with liquid isopropyl alcohol. And as usual, it is reacting with oxygen gas and we&#x27;re forming liquid water in carbon dioxide. Guess since we&#x27;re dealing with a more complex hydrocarbon here, I went ahead and wrote out the amount of each element we have on both sides to help us keep track. And as we did in the last example that we dealt with, we can go ahead and add our coefficient of too in front of our hydrocarbon. So based on that, we now have six carbon 16 hydrogen, and we now have four oxygen. From there, we can recognize that we can multiply our carbon dioxide by a coefficient of six to balance our carbon here. So we now have six carbon and we have 13 oxygen. And we now recognize that we can also balance or hydrogen from the left the rake inside by giving our water here a coefficient of Eygpt. So we now have our 16 hydrogen and we have 20 oxygen&#x27;s. Now, looking at our oxygen on the left hand side, we can recognize that there are two oxygen that are isolated and are hydrocarbon here. So we&#x27;re actually looking for what a factor we can multiplied by for our oxygen gas that will give us 18 oxygen from this compound so we can multiply by a coefficient of nine. So we have 18 oxygen here and two from our hydrocarbon, so that gives us a grand total of 20 oxygen and we&#x27;re now balanced. In this final reaction, we are dealing with solid lactic acid and as always, it is reacting with our oxygen gas and we&#x27;re forming the products, the liquid water and carbon dioxide guests. No, unlike the other two examples that we&#x27;ve done in this problem, Despite having this more complex hydrocarbon, we don&#x27;t actually need to add a coefficient of to hear in order to balance. We can start by recognizing our imbalance of carbon from the left to the right hand side. And we can add a coefficient of three tour carbon dioxide. Guess so. That now gives us at that lensed amount of carbon and an imbalanced amount of our oxygen. We can also look at our imbalance of our hydrogen here and recognize that we can add a coefficient of three in front of our water to balance. So we know have a balance number of hydrogen and we have nine oxygen on the right hand side so we can recognize that we have three oxygen that are isolated here in our hydrocarbon. So we&#x27;re actually looking for a factor that we can multiply our oxygen gas by to give us an overall number of six oxygen there being contributed from this compound here. So we know that if we add a coefficient of three in front of our oxygen&#x27;s, we achieve a grand total of nine oxygen on the left hand side and we&#x27;re now balanced

Aadit Sharma · Answer

so we have to re creation provided in the question. We just have to balance them. So for part of one, the balance chemical equation will be c two x 60 last three or do which leads to tow c 02 plus three water. For the second part, the balance chemical equation can be represented. Us, too. C 19 it Cuddy eight or do less 55 would do, which gives out 38 sea or two less 38 edge, too.

Problem

Explain why the combustion of glucose provides a …

Problem 5 Easy Difficulty

Write balanced chemical equations for the combustion of the fuels in the heats of combustion table on page 525. What generalization can you make about the amount of oxygen required for combustion of hydrocarbons?

Answer

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