Complete and balance the following molecular equations, and then write the net ionic equation for each: (b) Cu1OH221s2 + HClO41aq2¡ Write the net ionic equation for it.

Video transcript

Hello everyone. So in this video, we're gonna go ahead and find the net ionic equation for this given reaction. So first we're going to go ahead and write out the double displacement reaction by writing the products and because it's a double displacement, we're going to have ZN S. 03 and H 20. So by knowing our Celje bility rules are first starting material, these E N. O H two that's going to be a solid. Our H two S. 03 that's going to be a quiz. And then the product side we have these e. n. s. 03. Again, knowing our scalability rules, that's a solid and HBO is going to be water. So there's actually three steps of doing the net ionic equation first is by finding out the balanced chemical reaction. So we already wrote out our chemical reaction, but we have to make sure that it's balanced. So again, step one would be our mm balanced chemical reaction. All right. So on the product side and the reacting side, we see that we have CN oxygen. We have hydrogen and we also have sulfur of course. Same thing on the other side. On the product side, we have Zn oh H and S. Adams. All right. So on the left side we have one sink, five oxygen's 4 high gens and one sulfur On the right, we have one sink for oxygen's to hire a gens and one sulfur. How we can balance this is let's see if we put, let's see here. So we need This to be five oxygen's and this to be four high jeans, the only source of hydrogen is going to be from this right here. So what we can do is let's go ahead and double the Hodgins. So we put it to here. Then now we will have four hae jin's and five oxygen's total balancing out the left and right side. Alright, let's go ahead and also highlight that because that is going to be our balanced chemical reaction. All right, now moving on to our second step, our second step is going to be our ionic equation. All right, So what are ionic equation? Is is that anything that's not a solid or liquid from our balanced chemical reaction? It's going to go ahead and dissociate into its ionic form. So, first starting material that we have is that solid? So that will stay as is So we have a Z end oh H two, a solid Next starting material of H two s. 03. And that's so that will go ahead and associate we're going to have two H plus cat ions again, that is and then we're going to have one S. 03 to play atomic ion which is a risk as well for our products. We have the solid. So that should state as is and then we also have the H20 liquid. So that will also stay as is Alright, so that is going to be our ionic equation. Let's go and highlight that as well Now. 3rd and final step will be our net ionic equation. So Number three here is going to be our end E. So net ionic equation. So how we find this is by limiting all spectator ions. That's going to be all the ions that appear from the From step two are ionic equation. So we need to remove any ions repeat on the right side and the left side so we can see her that we don't have any ions on the product side. Of course we cannot cancel anything from the left side are starting materials. So actually our net ionic equation is going to be same as the ionic equation and just formally rewriting this. So we have our Z N oh H two which is a solid, we have our two H plus is which is a crisis. We have our S. 032 minus again, that is and that's going to make Z. N. S. 03, which is a solid And R two H 20. And that is a liquid. And this is going to be our final net ionic equation for this problem. Thank you all so much for watching