A student combines a solution of NaCl(aq) with a solution of AgNO 3 (aq), and a precipitate forms.

Hello. Everyone in this video we're trying to find the net ionic reaction for this reaction every year. So what we're looking at is going to be a double displacement reaction. We can see here of course we have a positive so I can go on and on on pair. What we're gonna do is just simply switch our partners. So if we're dealing with a double displacement reaction then so I'll just rewrite this reaction. So we have our F E. B. R. Two and K. Two S. And if we're going to switch this partner, so we'll just have our first Karen being F. E. And then our switching over partner instead of R. B. R. Two will be our sulfur. So us so that's going to be a solid. And then my next one. Let's see. So we have our K. And then we'll have our B. R. As a new partner and this is going to be a quiz. All right so let's go ahead and balance this reaction now and actually rewrite this because we need a little more space for the potential coefficients. So we'll have our F. E. B. R. Two. We have our K. To us. We have our F. E. S. And then we have our K. B. R. Which is acquis Alright, Sandra a line down the middle of the arrow to separate our star materials from our products. So we have F E B R K. And s. Of course. Same thing on the other side. Let me just extend this line. Okay so F E. B. R. K. And us. Alright so starting off we have one atom of F. E. Two atoms of br two atoms of K. M, one atom of our sulfur. And on the right side we have 111 and one. Alright, so to balance this out we see we just need our K and B are perfect. We have just that so we're simply add a coefficient of two if we do. So we see we have now two atoms of br and two atoms of potassium. Alright, so now using this bounced out chemical reaction, we can run our net ionic equation. So what a net ionic equation is is for any of our acquis um molecules it can turn into its ionic forms when it's in a solution and anything that's a liquid or solid will just remain as is it will not associate. Alright, so now we're writing our ionic equation. Okay, so we have first our first time material breaking apart into two F. E two plus and two moles of BR R. K. Two S will also dissociate into two moles of K plus And one more of us to -. Alright, so now we're dealing with our product side. So we said any solid or liquid will just remain as is so R. F. E. S will remain as is that KBR will dissociate. So we have two moles of K plus here And then we have two moles of RBR -. All right, let's stroll down to have more space. Alright, so now we're going to write out the net ionic equation which is just simplified to N. E. So, how we find the net ionic equation is by simply removing the spectator ions. So that's just ions that repeat on both the reacting side and the product side. So let's actually go ahead to rewrite our ionic equation. So we still see the step by step that we did clearly. So again, I'm just rewriting our ionic equation. Alright, so this victory ions that I see is going to boss first to be our And there are two K plus. That seems to be about it. So if we do that, then our net ionic equation is going to be FE two plus With their S two to give us f. E. S in its solid state. So this right here is going to be my net ionic equation for this double displacement reaction. Alright guys, thank you all so much for watching