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A metal $(M)$ produces a gas $(N)$ on reaction with alkalies like $mathrm{NaOH}$ and $mathrm{KOH}$. Same gas is produced when the metal reacts with dilute sulphuric...

Question

A metal $(M)$ produces a gas $(N)$ on reaction with alkalies like $mathrm{NaOH}$ and $mathrm{KOH}$. Same gas is produced when the metal reacts with dilute sulphuric acid Gas $(N)$ reacts with another toxic gas $(P)$ to form methanol at high temperature and pressure. ( $N$ ) also reacts with metals like $(Q)$ to form electrovalent hydrides. $M, N, P$ and $Q$ respectively are(a) $mathrm{Zn}, mathrm{H}_{2}, mathrm{CO}, mathrm{Na}$(b) $mathrm{Na}, mathrm{H}_{2}, mathrm{Cl}_{2}, mathrm{Ca}$(c) $mathr

A metal $(M)$ produces a gas $(N)$ on reaction with alkalies like $mathrm{NaOH}$ and $mathrm{KOH}$. Same gas is produced when the metal reacts with dilute sulphuric acid Gas $(N)$ reacts with another toxic gas $(P)$ to form methanol at high temperature and pressure. ( $N$ ) also reacts with metals like $(Q)$ to form electrovalent hydrides. $M, N, P$ and $Q$ respectively are (a) $mathrm{Zn}, mathrm{H}_{2}, mathrm{CO}, mathrm{Na}$ (b) $mathrm{Na}, mathrm{H}_{2}, mathrm{Cl}_{2}, mathrm{Ca}$ (c) $mathrm{Al}, mathrm{H}_{2}, mathrm{H}_{2} mathrm{~S}, mathrm{~B}$ (d) $mathrm{Mg}, mathrm{H}_{2}, mathrm{NO}_{2}, mathrm{Al}$



Answers

Balance each chemical equation.
a. $\mathrm{Na}_{2} \mathrm{S}(a q)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{NaNO}_{3}(a q)+\mathrm{CuS}(s)$
b. $\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)$
c. $\operatorname{HCl}(a q)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$
$\mathrm{d} . \mathrm{FeS}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{S}(g)$

All right, So we have more of these. Predict the products, balance the equation. Reactions. Right. So this is what it looks like a classic combustion reaction. So we're gonna predict the products here. Um, so it looks like we're burning ethanol, which is C two h 50 h. Otherwise don't have C two h 60 right. And because it's combustion, we're gonna have to go to and water as the products. And then on we have to do is balance, so right. But we could start with two carbon dioxides, So two carbons are all balanced now, um, we have three. Sorry, we have six. Hydrogen is on the left. We only have two on the right. We could put a three here to get those balanced, and then oxygen's here is what? It's gonna be a problem. Right? So if we count the number of options, it looks like we have three on the left right now. And after balancing the carbon and hydrogen, we have four from seal to. We're gonna have seven from water. So we're gonna have to do some balancing here. Right? Um So how do we get, uh, seven Oxygen's on the left. Um right. So because we have an oxygen in all of our products, we could do some simple algebra. I think I think this might be the best way to solve these problems. Right. So between the 1st 2 reactant ce, we have one oxygen and the first reacted, and we have two in the second, right? So, um, is there a way that we can get Ah ah coefficient in this position? That would have, um so call that X right? Plus the one that I'm talking about. This one is gonna be talking about that oxygen there. Um, and we're gonna have to have that equal seven. Right? And they're totally is right. So if we subtract one, we could get X equal to six. Um, so if we can get this whole term to equal to six oxygen, um, we'll have enough on both sides. And there's a really easy way we could do that. And that is when you get rid of this arrow, make this incredibly anticlimactic, but putting a three here, right? So then we're going to get six oxygen's from the 302 We're gonna get one from the ethanol sorry, I put zero. Really get one from the ethanol and six plus one nickel seven. So therefore, we could balance the equation. Um, by having two carbons on both sides, we have six hundreds of both sides, and then we have seven options. Um, on both sides as well. So that's a combustion reaction. Right? So here we have nitric acid and lithium hydroxide. So this is considered a double replacement reaction. Right? So, um, these are both Equus and Equus. So in order for these reactions to happen, um, you need to ah, have won your products be either insoluble a gas or a monk. Your compound. So it looks like as a result, you're gonna form lithium nitrate, right? And according to the scalability rules that Aaron textbook all Group one compounds are soluble and all nitrate disciple. So lithium. Um, with you, maitre, it's totally Sybil in water on. Then you get water, right? So you get otherwise in some of these reactions, you might see them as h o. H. Because you have the atrium nitric acid, reacting with the wage from the lithium hydroxide. Um right. And this is a liquid water is a liquid. So this reaction is going to go forward? This reaction does happen because we form a molecular compounds. Um, which is water in this case, Um, so that's a double displacement reaction. And then you get this reaction here. Right? So this looks like a double displacement reaction. So let's write the products first, and then we'll figure out if this reaction could happen. So you're going to get ah, lead carbonate when you swept so off the metals, Um, and then you're gonna get to equivalents of sodium nitrate. So again, with sodium nitrate, all group one and all nitrates are soluble is lead carbonate Sybil, though you know that lead could be an exception, um, to the rule. But the rule with carbonates is all carbonates except those of group one and ammonium are insoluble. So all carbonates are inside, will accept group one, and ammonia lead is not either of those. So lead carbonate. It's gonna be a solid. This reaction is gonna happen, right? So with these, I think the Net ionic equations are also really important. So the reaction that's happening here that makes this reaction actually happen, um is that you have lead and you have the car. The car box. That's hard box at their carbonate. Diana, I'm Polly Atomic eye on forming lead carbonate, which is a solid in water.

Problem. 1 12 Chapter three and chemistry Molecular approach. It wants to balance each equation. So So we go and a tu es es que plus see you and 03 to make you deals. And a and a three a cue C u s something. And an end to H four liquid, you and h three. Then we got h steal a cue. So you plus oh, to gas. Yeah, Ish to liquid. Plus, if yes, Okay, you h two s a a cue. All right, Now let's I'm sorry. It's due West Gas. So now we're gonna bounce it. We get to sodium and left, two on the two on the right. We have two nitrate on the on the right, two on the left, one cup on the left, one on the right. And when cell phone left, one on the right. That's balanced. Next be a tonight show on the left. We have three on the right. So And we have as and you'll see that we have 400 left in three on the right. So we got to find the least kind multiple of four and 24 3 So we're gonna have to put three here and for here. So we have 600 left. So sorry, we have 1200 left that well on the right. We have 806 100 left. We have 400 here and to here. So this is balanced. So next year we have 100 here to here. So get that too. Here. You have to chlorine on this side two on this side. And we want we have two options here, but only one here, So we're gonna have to change. Yeah, You're gonna to court two waters here is that we have two options here. Get to hear. Now we have 400 ends, so get that four out of here and we have four chlorine. So we have to have two in front of here. Right? Next one. I am Have one on the left. One on the right. Yeah, to corn on the right, one on the left. So we had to in front of here. We have two hours left to right, and we have one soul from left, one on the right, and that's what we do. We just do it step by

We have the reaction of ammonia week based with her court gases, which before we have to remember that money has a weak case, can donate. Electron. Is this if you were gonna donate electrons to this kind of my own Because clark acid actually associates were going to say to each plus feel of farmland, really get the nation. Is this electron to the hydrogen ion? And what we will be left with spinach for plus and feel? Oh, for minors except notice. Now that he's our guy on species, we're simply going to write some as so image for feel of war. Call that a quiz. These air office of a quiz. And thus my reaction when I write it cleanly is in a tree. A quid, a spill of four. Also a Prius. You know G makes the salt. I think so. Next in part B, we have the reaction between supercar greats, nitric acid. We have to remember that the reactions carbonates with assets, forms, carbon dioxide and water and the way we can figure that out is essentially do the double displacement reaction. Remember, this is a plus, Theo three to minus each plus and streamlining the double displacement reaction will quote unquote swap partners make compounds with following substances. But this will make his silver nitrate him the carbonic passion. Remember that curb on the cash in will decompose into water and Cartman accidents so that when we draw right the whole reaction we have a reactive just like me. A quick solution or solid. All right, A quick for now nitric acid sequence. He will form the salt silver nitrate. If you see a volleyball, this all nitrates are soluble. He will form liquid water and a gaseous carbon. Now it's reaction is not that I don't need to hear front of the silver nitrate and two in front of the nitric acid so highly important scene. Have the reaction strong. Think Lauren with water, Strong court quickly you water It's liquid. When a strong terms hey, light reacts with water when your atrocities it will make a strong team hydroxide trusses! We're gonna make strutting hydroxide. The oxidation state of Shelton will not change and our byproduct will be hydrochloric acid. This is essentially double displacement reaction again. Show that to you a little bit worse. Start plus minus treat water as H plus int o h minus. I think you will see that strong team hydroxide forms from the U hydrochloric acid through this.

All right. So here we have a question that's kind of four parts and asks us to take the element potassium and react it with kind of each of these elements rotten equation and then balance it at the end. So what we'll do is we'll just start with the first part and then kind of work our way on through until we finish. All right, so this 1st 4 is asking us to take potassium, which is Element K, and react it with water. Now here kind of just based on the form of the reaction, anytime you have a metal eye on reacting with a compound that typically is a single replacement reaction where the metal will be trying to replace one of the elements in the compound when the reaction is occurring with water. Typically, the metal is wanting to replace hydrogen. None of you. This is a little easier to rewrite water as h O. H. That way we can see that the potassium will try to replace the hydrogen, and therefore it will be making a new compound with the age that's left over. All right, But now the catches, any time you're doing a single replacement reaction. It's important to first check and make sure that this replacement can actually occur. The way we did this is by looking at what's called an activity Siris. What activity Siri's does is it list the elements in order of most reactive at the top two least reactive at the bottom. And so what we have to check is we need to make sure that the element by itself in this case, potassium, is higher than the element it's trying to replace, which in this case, is hydrogen. So if we go to our activity Siri's, which can typically find in the reference section of your textbook, or maybe in a handout that your professor is given you, we should check and make sure that potassium is higher up than hydrogen. It ISS so this reactions can occur. That's what will happen is the hydrogen that's being our place. We'll get kicked out by itself. Now. What we have to remember is when hydrogen is by itself, it's not just the element symbol age, but rather it's a TSH to that's because hydrogen and nature is found as a dye atomic element. Then what's going to be left over is we're going to have potassium now, making a brand new compound with the O. H I own left over from the water. It's old news will note kind of the charges of each eye on potassium is a +10 h is a minus one, and we'll balance those to create our new product. So it'll be k o. H as our new product. Then we'll go back and add in some coefficients to balance of the end. All right, so started Potassium. I've got one potassium on the left, one potassium on the right. So there should be good to go. Now let's take a look at our hydrogen Sze, I've got to Hodgins on the left, but careful. I've got 123 on the right. Okay, so no kind of that. The number two how the next biggest number it can go into. It's the number four. So I'm gonna see if I can make it where I have four Hodgins on the ride. If I had to hear, that will get me two plus two Hydrogen sze, which is for so I can go back and out of two on the left. So now I should be at a total of four Hodgins on each side. And then I think that will take care of my oxygen's to Oxygen's on the left to oxygen's are fried. And then finally, I'll go back and re balance the potassium by putting a two on the left as well. All right. And that should get us on that single replacement reaction between potassium and water. Okay, so let's go ahead and move on to the next part of this problem. Part B. So here the reaction is going to be very similar. It's potassium being reacted with a compound. So again, we're going to have a single replacement reaction where potassium will be trying to replace the hydrogen here. So again, any time you have a single replacement reaction, we need to go to our activity. Siri's and check to make sure that our metals potassium is higher or more reactive than the non metal is trying to replace, which in this case, is hydrogen. All right, so we'll have hydrogen. It's getting replaced. It will come out by itself again. Remember, when hydrogen is by itself, it's not just a tch, but H two since hydrogen is one of our die atomic elements. All right, so then finally well, old is will have the k left over, and it will be paired with the N H two that remains from the first compound. Those will combine in a 1 to 1 ratio to balance out the charges. And then last but not least, we'll go back and balance using our coefficients. So we have one potassium on each side. We have one nitrogen on each site. But now I need to do a little bit of work on the Hodgins at three. Hydrogen is on the left and then total on the ride. I have 123 400 ones. Okay, so I know that the number three how the next biggest number, it could go into a six. So if I put a two on the left, that would get me to six hydrogen sze, Then what I can go back and do is if I put a two on the right, I'll now have a total of 1234 Hodgins in the second compound. So therefore, Hodgins here, compared with two hydrogen is here. And so that will also get me a total of six Hodgins on the ride, then last but not least, I'll go back and check potassium and nitrogen to make sure that they stayed balance. So nitrogen looks good. I've got to know actions on the left to nationals on the ripe. But I now need to go back and add a two to potassium, since I added it to its A coefficient on the right. All right, so there we have it. So we have potassium again, participating in another single replacement reaction. Now for these last two problems, they're going to be a little different and that potassium is no actually reacting with a compound, But just another simple element. Some parts. See, we have potassium k reacting with Romain, which is B R. Two again. The reason it's not just your his bro mean is one of those special dye atomic elements that comes as a pair of Adam's whenever it's by it. So So this case, the type of reaction that we're looking at, is called a synthesis reaction, and since this is what we'll do is, we'll take the elements and will combine them to make one brand new compounds all right, so to make a brand new compound, it's important that we balance out the charges since this compound will be Ionic. So we'll take potassium and we'll also take romaine. And we're kind of right there. Charges over the top. If we look at our periodic table here, potassium is in group one, so it'll have a plus one charge and brew me is in group 17 or one away from a noble gas. It'll have a negative one charge. Okay, so those charges already balance out. So I'll just use one of each Adam to make the compound. Last but not least, I'll go back in balance by changing coefficients at one potassium on the left, one potassium on the right. That looks good to go to Broome means on the left. So come in and out of two on the ride. Then I'll go back and re balance. The potassium have to. Potassium is on the right, so go back and add a coefficient of two on the left. So that should have us for potassium plus priming yields, potassium bromine. And so again, just right quick, let me point out kind of a common mistake on these synths problems instead of coming to the side and balancing the charges like we did. It's kind of typical for students. Sometimes when they're learning these two just coffee, the K and the B R two from the other side. But that doesn't work, because here, bro Ming, is by itself. But here it's in a compound. So any time you're making a compound, don't just coffee the numbers from the left, but coming to the side, actually figure out the charges and then kind of re balance your compounds from scratch, that we can make sure that you're creating kind of a rial, you know, substance, huh? Then finally, let's tackle that last part of the problem where we're going to react. Potassium with Austin, it's again okay for potassium and then Yep, you guessed it. Oxygen is not just oh, but it's okay to again, since it's a dye atomic molecule. All right? And if you need a list of the dye Atomics right quick, just one thinking about it. There's kind of a handy acronym to remember which ones are in pairs of two, and they're single. Acronym is I bring cookies for our new home. And so that's iodine, bro. Me, chlorine flooring, oxygen, nitrogen and hydrogen. And so these are your di atomic elements. Okay, it's just in case you need that kind of a handy way to remember which one's bull into that category. Alright, Spectra problem suit Got potassium plus O two for oxygen. Again. What will do is come to the side and we'll balance our brain new product from scratch again. This is synthesis, so I'm taking both elements and combined them to form one new product. Potassium is a plus one. Oxygen is minus two. So in terms of balancing, this means that I'll need to Potassium is to combine with my one Austrian Adam. Last but not least, I'll go back and balance several equation by adding in some coefficients. So I have one potassium on the left, have to Potassium is on the right, so go back and add it to for my potassium on the left. Then I'll go on balance. My oxygen's up two on the left, so I need to put a two on the right. Then they have unbalances. My potassium is I now have two times two is four. Potassium is on the right. So I need to go back and change that first coefficient to a four on the left and again when you're working. These problems, it's handed have, um


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