What is the net ionic equation of the reaction of MgSO₄

Question

What is the net ionic equation of the reaction of MgSO₄ with Sr(NO₃)₂?

Accepted Answer

Hello everyone. So in this video we're trying to find a net ionic equation if it occurs. So the ration is when our sodium sulfate reacts with copper opacity. Let's find this reaction. Then so again, we have our sodium sulfate, which is just an A. Two S. 04 which is acquis that reacts with copper acetate, which is see you CH three c. O. And that is a curious so we see here right off the bat that we have a double displacement reaction. So, in a molecule of course we always have our catatonic and an ionic pairs. So we have our positive and or negative and positive and are negative. So what this means is that for double displacement then we're just simply switching our partners. So, but the first product then we will have our N. A. C. H three C. O. And their second product will be C. U. S. 04. Alright, so now we can go ahead to balance out our reaction. So, I put a line down the middle to separate our star materials from our products. She's lengthen this line. All right. So, for both sides, of course, we have our N. A. We have sulfur. We have oxygen. We have copper, carbon and of course our hydrogen as well. Same thing for the product side. Alright, this is just right off the bat. We have two atoms of sodium, one adam, sulfur eight of oxygen, one of copper and four carbon and six hydrogen for our product side. We have one of sodium, one of sulfur, six of oxygen, one of carper, two of carbon and three of hydrogen. So how can balance the right with the left side? Let's go ahead and add a coefficient of two to our first product. This then will give us to sodium eight total oxygen's four carbons and six Hira jeans. And now we can see that both sides is balanced. Alright, so with this balance can correction began to invite our ionic equation. Let's just scroll down for more space. So where ionic reaction, our equation is is just whatever that is. Aquarius from our chemical reaction when it goes into a quick solution to go and ionized itself and any solids and the goods will just remain as is. But in this problem we don't have that. So it was so basically everything will go ahead and associate into its ionic form. So for our first product again we're doing our ionic reaction now. Alright, so for our first product are first starting material that turned to two moles of R. N. A. Plus And then s. 0. 4 2 -. Then for our second reagent here to see you two plus and two moles of C. H three C. O. And then for our first product we will turn into two moles of an A plus and two moles of C H three, C 00 minus. And lastly our last free agent here we have c. U. two plus And s. 0. 4 2 -. Alright, now, finding our net ionic equation. So net ionic equation is by simply just removing Arnie Spectator ions and that's just ions that appear on both the right side and left side of the reaction. So what we do is first rewrite our reaction and while I'm doing this is so we can still see clearly our step by step. Of course we don't have to do this once we get paying of these. But for clarity sake or just Yeah. Alright, so we have officially rend our rear end the ionic reaction. Let's go ahead and take the time now to remove any secular ions. So we can see that the two moles of any possible cancel one more of S. 042 minus four. Council, I would want more of C. U. Two plus more cancel our two moles of CH three C O minus will also cancel. This means that we have no net ionic reaction because all the ions is actually moved. So we take a look now at the four different statements, that one that fits our reaction here would be deep. That there there's no net ionic vision that occurs. So, statement D is going to be my final answer for this problem. Thank you for watching

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