Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. c. Fe(NO3)2(𝑎𝑞)+KOH(𝑎𝑞)⟶

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Hello, everyone in this video, we're doing two parts. The first part is running the balanced net ionic equation for this specific reaction and part two, we're finding out what the spectator ions are. So for the first part fighting the net Ionic equation, there's actually three steps involved. First step is by writing out our balanced chemical equation. Second is writing out the ion equation and lastly we find the net ionic equation. So like I said, first step is going to be the balanced reaction. Alright, and like the problem said, these are the starting materials, so I just basically rewrite those. So we have our N. H four B. R. What is going to be acquis and then we're reacting that with a G. N. 03. We're just also going to be a chris And we're doing a double displacement formula here. So because of that on a product side we will have the NH four And '03 and knowing our saal ability rules, we know that this is going to be igneous and the second product is going to be a G B R, which is a solid. Alright now we have to make sure that this is balanced, how to do this is kind of do a telemark format of each pol atomic ion or element that's found in both sides. So we can see here that we have N. H four being involved, we have B. R. We have a G And o. n. three and just writing a line so I can separate the product side from the same material side. The same exact um team players here at age four B R AGNN three. Alright let's go ahead and count. So on the left side here I see that we have one NH four, we have one beer, one Kg and one n. 03 on the right side here we also have one and age 41 B. R one A G. And one and 03. So I can see from this telemark system here that both sides have the same amount of each element. And because of that this is a balanced chemical equation, let's go ahead and highlight that as our answer as well. Or the first step. Okay, so second step as I have mentioned, we're going to find the ionic equation. Alright, so finding the ionic equation basically from the first step. So what is highlighted here, we're just going to go ahead and change all the compounds that are not solids or liquids into its ionic forms. So basically you can see here in this whole equation that we have everything being a quiz. So those will dissociate into its ionic forms except for our solid here. So let's go ahead and do that. So for the first NH four BR compound it would associate into will have NH four plus that being acquis and then also be R minus which is also Aquarius. And then for the second starting material for the A G N. 03 that would associate into a G plus, which is a chris as well as en 3 -, also being acquis and then on the product side for the first Products of the NH four, No 3 Thou dissociate into NH four plus, which is egregious As well as N 3 -. Also Aquarius and the second product as you can see it's a solid, so we'll just go ahead and copy that down because that does not associate Alright, and that is going to be our ionic equation Alrighty. And last part and final part into getting our answer is going to be running out our net ionic equation. So how we do this is by simply limiting all spectator ions what are spectator ions and that's just any ion from part two here that we can see that repeats on both the product side and the starting material side. So if you can see from here, let's go ahead and actually rewrite that. So we don't go ahead and mess up our low system here. So we have the NH four plus, that is a quiz. We have our B R minus which is also Acquis, we have our A G plus which is a quiz, Then the n. also this And on the product side again we have NH four plus being Equus, We have our n. also being a gris and the second product here will not dissociate, so just be a G B R A solid. Alright, so if you can take a look and see which ions will appear on both sides, we can see here That we have NH four appearing on both sides as well as N. 03 minus appearing on both sides. So we eliminate those. Our final answer. Let's go and scroll down. Our final answer is going to be B R minus being acquis plus A G Plus, which is also Ignace and that will make a G. B. R A solid and that's going to be the final answer for part one Now, part two is actually really simple since we did part one already part to ask for the spectator ions and these bacteria ions are what we just crossed out. So we just put for part B here that the spectator ions is going to be an age four plus. Of course that's acquis And we also have or let's just put and 03 oh three minus being with And those are the two spectator ions here that we have from the first part, it's going to also highlight that. So we have here now the final answer for part one And we will also have the final answer for part two. Alright, thank you all so much for watching

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. c. Fe(NO3)2(𝑎𝑞)+KOH(𝑎𝑞)⟶

Verified Solution

Key Concepts

Net Ionic Equations

Spectator Ions

Solubility Rules