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CHEMICAL REACTICNSBalancing chemical equations with Interfering coefficientsBalance the chemical equation below using the smallest possible whole number stoichiomet...

Question

CHEMICAL REACTICNSBalancing chemical equations with Interfering coefficientsBalance the chemical equation below using the smallest possible whole number stoichiometric coefficients __NH,E (g) + 0,(g) + CH, (g) HCN(aq) + #,o()0-0 Dp

CHEMICAL REACTICNS Balancing chemical equations with Interfering coefficients Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients __ NH,E (g) + 0,(g) + CH, (g) HCN(aq) + #,o() 0-0 Dp



Answers

Complete and balance each of the following chemical equations. a. $\mathrm{C}_{8} \mathrm{H}_{18}(l)+\mathrm{O}_{2}(g) \rightarrow$ b. $\mathrm{CH}_{3} \mathrm{Cl}(l)+\mathrm{Cl}_{2}(g) \rightarrow$ c. $\mathrm{CHCl}_{3}(l)+\mathrm{Cl}_{2}(g) \rightarrow$

You see here three unbalanced e questions that you want to ballots. So I bring them out with some space for my coefficients. And my first step is going to be too. Take stock of the atoms that exists on either side of my reaction arrow here. So starting with the 1st 1 my reactant. I see I have one barium, and now I reach these this little molecule and perent disease, which means that I have two of this entire molecule. So I have, um, to nitrogen is here, and I also have three times to six Oxygen's here. Now, I also have oxygen's on this side of the arrow in this molecule, uh, four over here. So I have a total of 10 oxidants and I have to. So Diem's one chromium and I did already account for those four. Oxygen's there now, on the product side, I have one barium here, one chromium here. I have four oxygen's here, but I also have three over here, so I have a total of seven Oxygen's. I have one sodium and I have one naturally now, typically, the first thing I would look for it is the most complicated molecule. But in this reaction, all our molecules air Prue t complicated I made up of Ah, a lot of molecules. So what I'm gonna dio is see what looks similar on either side of my reaction arrow. For example, this n 03 is featured as a no. Three still on both sides. Um, and I have to on my react inside. So I'm just gonna try and see what happens if I put it to in front of this molecule. If I have two of those and I think but that well, actually solve it. So I have one barium, one barium, two nitrogen to nitrogen have six plus 4 10 oxygen's. I have six plus Port 10 oxygen's. I have to. So Diem's too. So Diem's one chromium, one chromium and, uh, already counted for those four oxygen's there so that Teoh did it. I'm going to turn it green because it is correct. And this equation is balanced now for our second equation here. Skin starting by just counting up. How many Ott Adams on either side of my reaction? Ero I have two carbons. I have five hundreds here and one here. So six total hydrogen ins, and I have one oxygen here and to hear. So I have three total oxygen's now on the other side of Mary my products. I have one carbon. I have Teoh Oxygen's here and one here. So I have three Oxygen's and I have to Hodgins. Okay. So the first thing I noticed is that this complicated molecule where I'm going to start has six hydrogen ins, and on the other side, I only have two. So, uh, the first thing I'm gonna dries to just get to six hydrogen is by multiplying this by three. So now I have six high virgins, and I have three oxygen's here and ah, to hear. So that gives me a total of five oxygen's. So I'm going to change this three here to a five. No. Okay. Right. All right. So now I have more oxygen's on my product side, so let's see what I could do. So if I add if I wanted to make this oxygen, uh, equal toe three. So I have three plus two to get five. That changes a lot of other numbers here that gives me, you know, six carbons and fact. So I'm just going to see what happens if instead, I changed this, um, oxygen. Uh, so if I multiply this one by two, I have four oxygen's here and one here. That is fat. So let's take stock. I see two carbons here and Onley one over here. Okay, that's something to note. I have six Hodgins and six hydrogen engines. And, um, four post 15 oxygen's and three plus 25 oxidants. So what happens if I change this C 02 to give it a to coefficient so that I have two carbons on both sides, But that's also gonna change my number off Oxygen's on this side. So I'm precursor early, going to erase this to cause I don't think that's what we want there. So now I have two times 24 oxygen's plus three oxygen's. So I'm going to raise this five here, and I have seven over here now. Yeah. So what can I do to get seven auctions over here? I have one here, so I need this to equal six. I can do that by putting a three here, right? And I think that might have done it. Let's see, I have two carbons, two carbons. I have five plus 16 hydrogen ends. I have three times to six hydrogen ins. I have one plus three times to its 61 plus six is seven oxygen's and I have for oxygen's plus three is seven oxygen's. So now that all my numbers match, I can turn my coefficients green. This not a couple steps back, So I have one here. Don't need to write a three here. She who's your twos? And three waters on. And that is a balanced equation. Yeah, last one. Let's see. I can stock up on my Adams. I have one. Excuse me. Calcium. I have two carbons. I have to hide regions and one oxygen. Those are my reactions. On the product side, I have one calcium e. Have Remember this to outside the prince. Evening. That goes to both. I have. So I have to. Oxygen's here and to hydrants here. I also I'm seeing two. Hydrogen is over here, so I'm gonna make a note that I have a total four heart surgeons on this side and I have two carbons. So the first difference I noticed is that I have more hydrogen ins my products than I do in my actives. So I'm going to see if I can match number pathogens. Uh, the easiest way, uh, is that I'm gonna try first. So multiplying this h two molecule by to get a total four Hexion Adams. And that also means that I have two oxygen atoms now, and I think that might be it. One calcium, calcium, two carbons, two carbons, four hydrogen ins to and to four hydrogen to oxygen's and toe Oxygen's already changed this to two grand and have a balanced equation.

Okay, lets try balancing these chemical equations. So if the 1st 1 things to look for Well, we have the same number of calcium is on both sides. Good. Let's do oxygen's next. So I have one extra on the left. I have to auctions on the right. OK, so I can fix that by throwing a to in front of this. Okay, so that fixes The oxygen's down. Let's look at hydrogen. I have four on the left and I have four on the ranked. Um oh, excuse me. No, I actually So let me let me write this out. So I have four eyed virgins here. I have two. Hydrogen is here. This is two hydrogen. So I have a total of six on the left right now, I only have four until I want this to be four. And I can do that by sticking to outfront of here. So now everything is balanced. Okay, Uh, so for the next one, um, it's easiest if you condensed this a little bit. So let's let's ah, make this a little bit easier to see, and I'm gonna write this as c and there's four carbons. Ah h They're six. And then there's to be ours. And then on the right hand side, I'm right. This as C four h six b R four is the number of bro means. Okay, so the only thing I gotta watch out for is the bro mean, that's the only thing that's different from left to right from reacting to product. And I can fix it by sticking it to right there. Okay, then for the next one leads correct on both sides hydroxide. That's though I have to hydroxide on the left. Your sorry on the right. I have one on the left. I can fix it by throwing its you in front of that hydroxide. And now that equation is balanced for the bottom. Let's go ahead and to nitrogen is first. So one nitrogen on the left, one on the right. Good. Let's do hydrogen ins. I have two on the left, one on the right. That's not gonna work. So throw it to in front of here to get the hydrogen is correct. But by doing that, I've messed up. The nitrogen is right. So now I have three night regions on the right. I have only one on the left, so I'll stick a three out here. Let me try that again. It's like a three out here. That way I have three nitrogen on the left, Three on the right, and let's check the number of Oxygen's auctions. I have six here, one here. So I have seven on the left and I have six year one here. I have seven on the right.

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're reacting that with hydrogen gas and we'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'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'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're forming liquid water in carbon dioxide. Guess since we'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'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'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'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're forming the products, the liquid water and carbon dioxide guests. No, unlike the other two examples that we've done in this problem, Despite having this more complex hydrocarbon, we don'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'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's, we achieve a grand total of nine oxygen on the left hand side and we're now balanced

In this problem said, We're balancing combustion reactions. The problem specifies excess oxygen, so that means we're dealing with complete combustion on the products of a complete combustion reaction of a straight hydrocarbon like propylene. Here are CO two and H 20 so balancing these is usually pretty straightforward. We have three carbons on the left. That means we need three carbons on the right and we have six hydrogen zones. We need six. I'd regions two times three is six and now we basically look at however many oxygen's. We have a figure out what needs to go here. So here we have two times three is six and we have another three, so we have nine oxygen's. We can take care of that by putting a nine halves here and if we want the clear of the fractions, which we don't have to, but it often looks neater. If we do, we can multiply everything through by two. That will give us to probe Ulene's nine oxygen given six CEO, too, and six water. The next one is a little bit different because it's dial benzo a gas and it's got a sulfur in there. That means the products again. Complete combustion. Because we've specified access. So too. But the products are now CEO two and s 02 and water because the soldier has to go somewhere but doesn't change things very much. We have six carbon going in here, so we need six carbon. Uh, actually, you know what? That seven So seven carbons. And now we have 56 I'd regions. So we need sex. I'd regions, we have one sulfur we have once over. Okay, now, what about our oxygen? Here we have 14 and to and three. So we have a total of 17 oxygen's so we could right, 17 have. Except we have one oxygen there, so we only need 18. So that's knowing. And this next one, it's actually kind of helpful. This is literal, and the formulas written this way because it gives some indication of the structure. But for balancing it might help ASU. If we count, We have one, 23 three carbons and we have 23456 78 id origins and we have 123 oxygen's. So from the point of view of balancing, that's actually an easier way to look at it and begin complete combustion. Our products or co two and H 20 Okay, so three carbons, three carbons, eight hydrogen, eight hide regions. How many oxygen does that give us on the right? We have six. We have another four. That's 10 3 taken care of here. So we need another seven. We could make that a seven halves and again if we want to clear the fractions, but we don't have to, but sometimes it's regarded as your We can multiply everything by two. We're gonna


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