predict-the-products-for-the-reactions-given-below-be-sure-to-balance-each-equation-a-mathrmhfa-qm-2

Question

Predict the products for the reactions given below. Be sure to balance each equation.
a. $\mathrm{HF}(a q)+\mathrm{NaOH}(a q) \longrightarrow$
b. $\mathrm{HCl}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(a q) \longrightarrow$
c. $\mathrm{HF}(a q)+\mathrm{N} \mathrm{H}_{4} \mathrm{OH}(a q) \longrightarrow$
d. $\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow$
e. $\mathrm{HNO}_{3}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(a q) \longrightarrow$
f. $\mathrm{HNO}_{3}(a q)+\mathrm{NH}_{4} \mathrm{OH}(a q) \longrightarrow$
g. $\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(a q) \longrightarrow$
$\mathrm{h} . \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NH}_{4} \mathrm{OH}(a q) \longrightarrow$
i. $\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{HNO}_{3}(a q) \longrightarrow$
j. $\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(a q) \longrightarrow$

Vishal Sharma · Accepted Answer

in problem for predictive product and balance each off the following equation. So first days when hydrogen fluoride reacts with the sodium hydroxide so it will form sodium fluoride and h toe. So this is the product off reaction. A. In the reaction be and it is balanced equation in direction be when hydrogen chloride at sale reacts with the magnesium hydroxide so it will form magnesium chloride. That is mgl too, along with H two liquid. So on balancing this occasion, we get to excel. And now this is the balance equation for this reaction. In a reaction C when hydrogen fluoride reacts with the sword ammonium hydroxide that is energy for college. So it will form ammonium fluoride. That is energy for F, along with H 20 So this is again a balance A situation. So in the that is when ascetic acid ch three c double ohh reacts with the sodium hydroxide so it will form sodium hesitate, which is ch three c double o n e along with H tour liquid. So this is again balance situation. So in party when at Channel Three, that mean nitric acid reacts with the magnesium hydroxide, so it will form magnesium nitrate, that is M G N 03 twice and h toe on balancing. We get, uh, coefficient off to in a general three and two h two. So this is the balance equation now, in part F in part, off off the equation. So, in part of off the equation, that is the, uh when at Channel three reacts with the ammonium hydroxide ammonium hydroxide, that is energy four or age. So it will form ammonium nitrate, NH four and no. Three and water and water. So and this is again a balanced equation. This is again balanced equation now in part G that when ascetic acid reacts with the magnesium hydroxide So this is the ascetic acid. When it reacts with the magnesium hydroxide, that is M g o H toys. So it will form magnesium as it did that is M g ch three c double o twice less XTO plus edge toe. So on balancing we get here, we get to as it stock American vision is to, and this is to which tour. So this is the balance situation now in part edge, when ascetic acid reacts with the ammonium hydroxide. So when ascetic as it wrecks with the ammonium hydroxide NH four h So it will form. In this case, ammonium as it did that is ch three c double o ch. Three c double o NH four and H two. So this is again a balanced equation. Now, in part, I off the equation is when h two s 04 react with the channel three. So So it will form h s or four minus equals less and no two positive equals and h 30 positive backwards. So on balancing we get so we get this equation. So this is the balanced situation? No, in the part G. So when sulfuric acid that is h two s 04 Uh, this as it reacts with the magnesium hydroxide m g o h twice. So it will form magnesium self with that is MGs off four and h 20 So this is the balanced equation for this case. So So these are the balanced equation, so I box in green color the answers. Now these are the answers. So these are the in the right hand side, this show our products and in the left hand side, these are the reactant. So we find the products as well as we balance the equations. So these are over answers. Let me highlight so that this is our answer for this. These occasions can understand better for these equations like this.

David Collins · Answer

here. We have several chemical reactions. A little balance and identify. The 1st 1 is ammonia reacting with hydrochloric acid or HCL producing ammonium chloride. It&#x27;s already balanced. They have combined. So this is a combination reaction. But then have a hydrocarbon reacted with oxygen producing carbon dioxide and water. So this is a combustion reaction. This takes a little bit more to balance back up just a little bit. We&#x27;ll see that we have four carbons, so we&#x27;re gonna need to put a four right here. Now, um, we have eight hydrogen, so we need to get where you need to put a four right there to get eight hydrogen. Now, we have generated 12 oxygen, so we need to put a six right here to get our 12 oxygen&#x27;s for the next reaction. We just have antimony combining with chlorine. So this is a combination reaction. We have, um, six. Well, right now Or we could make the chlorine is equal. So we have three chlorine and two chlorine, so I need to put it to right here to get six chorines, and then we need to put three right there to get six chlorine that introduced to antimony. So we need to put it to right there for antimony. And this is a combination reaction. The next one. We have nitrogen tri iodide falling apart into nitrogen. And I we have two nitrogen, one nitrogen. So we need to put a two right there. That then gives us six. I dines, so we need to put a three right here to get six ia dines. For this is the decomposition reaction. Give er reacts with C l to producing K, C, l and B R. In order to balance it, we have to see l. So we need to put a to in front of the K C L. We have to be ours. So we need to put a to in front of the KBR. This would then be a single replacement reaction as chlorine replaces roaming. We have iron combining with sulfuric acid, nobody&#x27;s iron, three sulfate and hydrogen gas. This is a single replacement reaction where the iron replaces hydrogen. To balance this, we have three sulfates, so we need to put a three right here. We have two irons, so we need to put a two right there and then we have six hydrogen, so we need to put a three right there, and we have aluminum sulfate reacting with sodium hydroxide, producing sodium sulfate and aluminum hydroxide. This will be a double replacement reaction in order to balance it if we have to. So Diem&#x27;s here would have to put it to here. But then, that&#x27;s not gonna balance our hydroxide. We have three hydroxide, so the two of the three gives us six. So now we&#x27;ve got six sodium hydroxide, which will require a three to get our six so Diem&#x27;s and then to hear, to get our six hydroxide and then that will give us the two aluminum&#x27;s, which we have over here with our three sulfates, which we already have right here.

Matthew Lueckheide · Answer

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&#x27;re gonna predict the products here. Um, so it looks like we&#x27;re burning ethanol, which is C two h 50 h. Otherwise don&#x27;t have C two h 60 right. And because it&#x27;s combustion, we&#x27;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&#x27;s here is what? It&#x27;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&#x27;re gonna have seven from water. So we&#x27;re gonna have to do some balancing here. Right? Um So how do we get, uh, seven Oxygen&#x27;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&#x27;m talking about. This one is gonna be talking about that oxygen there. Um, and we&#x27;re gonna have to have that equal seven. Right? And they&#x27;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&#x27;ll have enough on both sides. And there&#x27;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&#x27;re going to get six oxygen&#x27;s from the 302 We&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s a double displacement reaction. And then you get this reaction here. Right? So this looks like a double displacement reaction. So let&#x27;s write the products first, and then we&#x27;ll figure out if this reaction could happen. So you&#x27;re going to get ah, lead carbonate when you swept so off the metals, Um, and then you&#x27;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&#x27;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&#x27;s happening here that makes this reaction actually happen, um is that you have lead and you have the car. The car box. That&#x27;s hard box at their carbonate. Diana, I&#x27;m Polly Atomic eye on forming lead carbonate, which is a solid in water.

Kaitlynn Wade · Answer

In the first example, we have two compounds in the reactant and two and the products. However one is missing. Already looking at this reaction, I can assume that it&#x27;s a double replacement reaction as we have two compounds and the reactions that can displace each other, which has seen in the first reaction that is already in the equation. As zinc displaces in H four. Two pair was so fired then we can assume that NH four will also display sink and bond with core. I&#x27;d therefore, the reaction. The missing product will be image for CEO and to balance his equation, we can easily just added to go fishing to which will double the amount of image foreign quarried in the products and therefore make equal on both sides. The next example. We have aluminum plus lead in in No. +32 Looking at this, we can assume that it&#x27;s a single displacement reaction. Since aluminum is by itself and PB and No. Three, there&#x27;s a compound and PB and aluminum are very similar and so aluminum can this place PB. Therefore, the products will be aluminum in 03 to cost PV and since aluminum and PB both have an oxidative state of plus to the other three will be balance on both sides as well as aluminum and PB. The next example we have is be a pose h 20 This also could be a single displacement reaction and that would mean that the products will be E nao plus age too. And since hydrogen will pair about itself than be alone, it will create an equal amount of Adams of both sides as well as the barium in oxygen. And this is also a single displacement reaction. The next reaction we have is sealed two plus KBR. This also is a single displacement reaction as chloride can replaced Romain and here with potassium. Therefore, the reaction will be K CEO close. They are too. In order to balance the equation, we can add a coefficient in front of the case CEO, which will give us an equal amount of Cory. And then we can also add a coefficient to KBR to balance out the potassium and grooming. The next example is in H three plus 02 and N o plus H 20 So this one, we don&#x27;t have to fill any missing products in, but we do have to define what basic reaction is saying and balance it. And so for this reaction, it is a combustion reaction. As we can see, it creates water and in zoos oxygen and to balance the equation, we can start off by balancing the amount of hard region, which, just for the amount of time I&#x27;m gonna go ahead and put a coefficient of six in front of H 20 And then that will allow us to have 12 hydrogen, which we can come over here and not a coefficient of four to an age three. To balance that out to get the number of nitrogen balance. Now we can add an another coefficient four to N. O, since four times one equals four and then to balance the oxygen, we see that we have a total of 10 in the reactant, and so we can easily at a coefficient of five in front of the 02 and the next example. We can see that it&#x27;s a decomposition reaction, since there&#x27;s no reactions or products that we need to fill in in a starts with a compound and breaks up into smaller ones. and in order to balance this equation will need to double the amount of oxygen in the reactor. Reactant so we can add a coefficient of two because two times one equals two. And that will also double the amount of heart surgeon in two times two equals four, So we cannot another coefficient of too in front.

Julie Guaglianone · Answer

problem. 40 asks us to balance these equations the 1st 1 for the 1st 1 We have to sodium on the left to need to on the right know how to chlorine on the right. You have to quarry on the left. Stats bounced the next one. We need a three overhears. We have three irons on each side. We have four Oxygen&#x27;s on the right. Who went out of four here to get four Oxygen&#x27;s in the left. And now that we have eight hydrogen zahn the left one out of four over here. So we get eight hundreds on the right to know that&#x27;s bounced for this one. You have to chlorine chlorine on the right to one out of two here. So we get to Koreans and left. And now we have four hyphens on the left. So we went out a two year. So you get four hydrogen zaun, the right that also balances are oxygen&#x27;s with two on each side for the next one. We have to bro means on the left. So we went out a two over years. We have two on the right, and now we have four hydrogen ins on the right. So we went out of two overhears. We get four hydrogen ins on the left, and now our oxygen&#x27;s are even with four on each side. So that equation is well balanced. This next one I need at a three over here. So we have even sodium on each side. We now have seven Oxygen&#x27;s on the left, and we have five on the rates. We want to add threes. We get four here, three here to make seven and know lots even. And then we also have balanced than our hydrogen ins with six on each side. That equation is now balanced in this one. We haven&#x27;t even number are unequal number of sodium and nitrogen ins on each side. And for the oxygen&#x27;s, we have three on the left and four on the right. So to balance this, we would really need 1/2 over here. But because you can&#x27;t have half coefficients in balanced equations, you need to add a to here. Mitu here as if you had multiplied all of them by two suiting up with the one in front of the ah co two. So for this next one, we want to look at the sodium you have to hear. You had to hear. So then, with the oxygen&#x27;s, we have three on this side and four on this side. So to balance that, we want to again. This is as if we have 1/2 here. So you have to multiply everything by two. So we have to to for and then because this was 1/2 it&#x27;s now one because he multiplied by two. So for this last one, you look at their eight sulfur Adams here on the left. So to make that even we wanna put to here Now we have spore silicone on the right. So when I have gore on the left and now that&#x27;s balanced.

Julie Guaglianone · Answer

problem. 39 asks us to balance these chemical equations for the 1st 1 We have to potassium on the left. We need to make to potassium on the right. So that balances our equation. The second equation, when iron the left when I earned the right two. Hydrogen is on the left to hydrogen on the right, one oxygen on the left, one oxygen in the right. This one is already balanced. And then, for this next one, we have one sodium on the left. One study on the right, when the Korean on the left, one corn on the right by oxygen&#x27;s in the last five. Oxygen&#x27;s on the right to hide regions on the left to hydrants on the right. This one&#x27;s already balance as well. This next one and we have to Rangan ease on the left. So balance needed to. Over here we have three oxygen&#x27;s on the lefts and add three here and now we need also add three on the left in front of magnesium to balance. So for this one, I looked at the number of high trojansfirst and I could see that there three on the left, She on the right so to make four on both sides are edited to here and to hear. And this also balances are potassium with three on each side, and our oxygen&#x27;s are also balanced. So this equation is balanced. So for this next one, yeah, the number of nice chins are equal on both sides. The number of hydrogen xyz to over here. So but it you over here and then Now you have to Nitrogen is on that side. You need to put two over here. But now the number of oxygen&#x27;s we have four, 567 and we have six on the sides of Really, if we could, we would want 1/2 here. But since we can know what we can&#x27;t have 1/2 as a coefficient, we actually need to multiply across bright too. So we end up with for one of that to their This is one and then this is for Is that its balance for this one? We have one barium on the last one. Bury him on the right. We have to hide regions on the right to hydrogen is on the left. We have three oxygen&#x27;s on the left and three oxygen&#x27;s are four Oxygen. I&#x27;m sorry. On the right. So, in order to balance this, I started by putting two here. So now we have four hydrogen ins. Now we have four Hodgins on both sides, and then we need to put it to over here. So, you know, but to burying on both sides and we end up with four plus two is six. I mean, a before plus two is six for Oxygen&#x27;s as well. In this last one, we have three hydrogen ins on the left and two headers on the right. So do fix that to over there and three over there. It&#x27;s now going to make sure that our nitrogen zehr balanced. And then if we counteractions, we have two on the left and five on the right. So to balance this, who would actually need to do five halves? And again? Since we can&#x27;t have fractions, we have to multiply by two through, So it should actually be four on the five and then four and six, and now it&#x27;s balanced

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Suppose you mix 1 mol of sulfuric acid, $\mathrm{…

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