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8.12 Balance each ol the Tollowing chemical eyuations: K(s) + HOm Hg} KOHtaq} Cnts) S (s) CrS;u} HO() H,BO;(aq) HClaq) BCI;G) H,O() Fe_(SOa)staq) 4.HSO_aq) Fe(OH)(s...

Question

8.12 Balance each ol the Tollowing chemical eyuations: K(s) + HOm Hg} KOHtaq} Cnts) S (s) CrS;u} HO() H,BO;(aq) HClaq) BCI;G) H,O() Fe_(SOa)staq) 4.HSO_aq) Fe(OH)(s) Bu,(PO Us) NClaq) BaCl,(aq) Na,PO_(aq)

8.12 Balance each ol the Tollowing chemical eyuations: K(s) + HOm Hg} KOHtaq} Cnts) S (s) CrS;u} HO() H,BO;(aq) HClaq) BCI;G) H,O() Fe_(SOa)staq) 4.HSO_aq) Fe(OH)(s) Bu,(PO Us) NClaq) BaCl,(aq) Na,PO_(aq)



Answers

Balance each of the following chemical equations by inspection: (a) $\mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{NH}_{4} \mathrm{OH}(a q) \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ (b) $\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\operatorname{NaBr}(a q) \rightarrow \mathrm{Hg}_{2} \mathrm{Br}_{2}(s)+\mathrm{NaNO}_{3}(a q)$ (c) $\mathrm{Mg}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \rightarrow \mathrm{Mg}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{H}_{2}(g)$ (d) $\mathrm{LiNO}_{3}(s) \rightarrow \mathrm{LiNO}_{2}(s)+\mathrm{O}_{2}(g)$ (e) $\mathrm{Pb}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{PbO}(s)$

Let's balance each of the following Chemical equations for E. A is balanced as is and doesn't need any added coefficients for B to balance, we need to put a two in front of the cobalt knight, right three in front of the ammonium sulfide and a six in front of the ammonium nitrate. For C to balance, we need to put a two in front of the lithium hydroxide on the product side and for D. We need to put a two in front of the K. C. L on the reacting side and two in front of the potassium acetate on the product side.

Okay, easy question. We just have to balance the given equation and sold in hydraulic plus water. Yeah, we have sodium hydroxide. That's hydrogen should be. We have iron plus H two was so full leads to formation of himself A. Sure. That's hiding okay. Yeah. Uh We have hydrogen and beryllium. We have to H. B. R. Option D will be sodium plus hydrogen to N A. Ahh. And lastly, we have led oxide plus hydrogen, which leads to formation of lead plus billings of water.

Okay, We have a few chemical reactions that we have to balance their currently unbalanced. Uh, so the first time I like to do is just to write out the equation with some space for my coefficient. So that's what these little like underlined sections are, Um, So the next step is to just get ISMs for which molecules we're looking at and how many of each data we have on either side of the reaction and rows that we have our reactions and our products. And here on erect inside, we have ah, one calcium. Our matter. No subscript needs one. Um, we have two florins, 200 ins, one so for and four oxygen's. Um, one way people do this is to also, like, ride out the, um, molecules under. So, you know, we have, uh, once he a Teoh, two ages, etcetera, and, uh, on the other side, the products we have one calcium one. So for for oxygen's one hydrogen and one flooring. And so now we can get a sense for the atoms that are on the even unbalanced. So we need more florins on the product side. We need more hydrogen Z on the product side, Uh, we're equal on the calcium were equal silver, and we're equal on oxygen. So we have 100 imploring together in this molecule here, and this is the one that we need more off. So since there are to hide regions and two florins Ah, let's throw it to here. 10. Let's just check our work. Whether they match. Okay, one can see what Calcium two florins to floor. Eanes 200 wins, two hydrogen engines. One sold for once over for oxygen's for Oscar. It's great. So I'm going to turn this to green. And that will be our final answer for this one. So could rewrite this. If the coefficient is one, you don't need to write anything, so you would just right cf two plus h two s 04 reacts to form CSO four plus two h. All right, you bought to the 2nd 1 So again, I'm gonna start with just taking stock of what we have on each side. I have one potassium, one, bro. Mean three hydrogen ins, one false for us and four oxygen's. Um, on the product side, I have three potassium, one phosphorous, four oxygen's one hydrogen and one bro. Mean so I know that I need more potassium seems on the react inside he and more hydrogen on the products. Uh, usually, you would start with the most complicated molecule. So I have on their react inside h three p 04 and our peas and owed are matching on both sides. But this hydrogen has more than the other side. So Ah, if I've three hundreds here, let's just see what happens if I make this equal by putting a three as coefficient. Now, if I do that, I now have three bro means on the product side and only one Berlin, the reactive side. But if I add a three here, let's see how that works. So now I have three pro means on each site. Looks good. Now I have three potassium on my react inside. And oh, I also have three potassium on the products that. So just gonna check a gun through potassium. Potassium is three, bro. Means three bro means three Hodgins to hide regions. One potassium, one potassium for oxygen. For Austrians. Great. So both sides are equal. And I'm going to change these threes to green because they are correct. And this is my final answer. Okay, let's try the next one. So again, stock. I have one Titanium at four. Corinne's one sodium. That's my react inside now, products. I have one sodium, one chlorine and Juan Titania. All right, so I'm looking at my more complicated molecules, which would be my, uh, t I seal four on my n A CEO. So I see that on one side, four chlorine. And now I need to see what happens. Let's just try four to get four chlorine on my product side as well. Now, if I do that, I now have four. So Diem's on my product side and only one on my rackets out. So, in order to get four, I need to add that coefficient four over here as well. So see what we're looking at. One titanium on the side, One titanium on this side four. Chlorine is over here for Koreans over here for so Diem's here. And four. So Diem's here, so that is correct. So I'm gonna change my force to green. Close. This is my final answer. And last but not least, here we have a gander, starting with my chicken stock I have to potassium atoms, one carbon atom and three oxygen atoms on the reactant side. On the product side, I have to potassium atoms. Here's one oxygen and there's two over here. So I have three oxygen atoms and I have one carbon atom. So let me just read. Check that to potassium zero to pass. Keeps here. One carbon here, one carbon here, three. Oxygen's here. Three Oxygen's here. This equation is already ballots. We do not need to add any coefficients here. Her So here are our final answers again or balanced invasions.

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.


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