Write balanced molecular and net ionic equations for the following reactions, and identify the gas formed in each: (a) solid cadmium sulfide reacts with an aqueous solution of sulfuric acid

Video transcript

Hello. Everyone in this video we're trying to find the net ionic equation for a specific reaction. In this case we're dealing here with P. D. O. N. H. N. 03. So there's actually three steps to do when we're doing a net Ionic equation first is to write out the balanced chemical reaction and then the ionic equation. And lastly the net ionic equation will be which will be our final answer. So like I said the first step is writing out our balanced chemical reaction. Alright so starting re agents like I said we have the P. D. O. Which is a solid that's going to be reacting with h. n. 0. 3 which is a quiz. We're doing a double displacement here. So in that case on the product side we will have P. d. n. 0. 3 2. That's going to be Aquarius. And we also will have water on our product side which is a liquid. Alright so how I want to go ahead and check if it's balances by kind of doing like a tally mark for all the atoms involved. So we have P. D. We have oh we have a church and we have N. Yeah I kind of write A or dry vertical line, just dividing the product side and starting reagent side. Alright, the same exact order. So palladium, oxygen, hydrogen and nitrogen. Alright so on the left side you can see we have one pt we have four Os. We have one H. And one end On the right side. So the product side we have one p. D. We have seven Os. We have two ages and we have two ends. So we need to balance this out and doing it in this format. I think it's a little bit more clear to see. Let's go ahead and manipulate the H. And the ends. And the only source of H. And ends will be this H. N. 03 here. We want to go ahead and change this from 1 to 2 so it can be balanced. So we'll go ahead and write in a. two in this case. Now we will have a total of 70. S. We'll have to H. Is and we will have two ends. So now it's going to be balanced because both sides have the same same amount of the atoms. So one P. D. Seven Os two H. S. And two ends. All right. So that's going to be our bounce. Chemical equations go ahead and highlight that as well. Alright and next we have our ionic equation. Let's go ahead and actually scroll down. So we have a little bit more space. Mhm. Alright this is good. Okay so for Ionic equation anything or everything but a liquid solid will dissociate into its ionic form. So from this equation right here let's go ahead and do that. So P. D. O. Our first Honoree agent that's actually solid so we don't touch that. So. Alright so PDO as a solid And then next we have two h. and threes and that's that is going to dissociate. So we will have two H. Plus and again that's a chris and we will also have two N. 03 minus Aquarius. So you can see here that because this increase and it's not a solid or liquid, it will associate into its ionic forms. In that case it would be a plus N. +03. Now continuing on to the product side, we're doing the same exact thing. So we will get P. D. Plus which is Aquarius, You have two and 3 again. That is a curious and then we have hte jewel which is a liquid. So we don't touch that. All right so that state didn't really fit in the frame but H. 20. Is a liquid. Alright perfect. So that's it, that's it for the ionic equation let's go and highlight that. Alright. And last step is the net ionic equation. So we'll abbreviate that as N. I. E. Alright, so how do we get to the net ionic equation? So we have to kind of look for our spectator ions and that is just ions that repeat on both the starting material side and the product side. And it doesn't really play much of a role in the chemical reaction. So we can eliminate that. So let's go ahead and rewrite what we have written on top which is just the Ionic equation. So that is P. G. O. Solid plus two age plus can actually scroll down a little bit so we can visualize this better. That's a curious then we have our two N. 03 minus again. That is a chris. And then on our product side we have our P. D. Plus which is Aquarius. We have our 2.03 - which is Aquarius. And of course we have that water and its liquid form. All right. So let's look for ions that repeat on both the product side and the starting rage inside. We can see here that the only one that repeats will be the two N. 03 minus. So we can go ahead and cross that out. And once we have crossed that out we'll go ahead and rewrite what is the only thing that is left? So we have our P. D. O. Solid Plus two H. Plus which is Aquarius. And because we eliminate the n. 0. 3, we go ahead and not include that. And then we have our P. D. Plus and then our water H. 20. Which is in this liquid form. So that what I have written in black is going to be our net ionic equation. And that is also our final answer for this problem. Thank you all so much for watching