Complete Ionic Equations Example 2

by Jules Bruno
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based on the given reactant provide both the molecular equation and the complete ionic equation. So here we have ammonium sulfate react with calcium chloride. So for steps zero, we're just going to follow steps 1 to 4 that we've learned in the past to first give the molecular equation. Now, remember, when it comes to these steps were first going to break up each of these compounds into their ionic forms. So ammonium sulfate breaks up into the ammonium ion and the sulfate ion. Don't worry about the fact that we have to Ammonium is here. We'll discuss that later on as we're going into the complete Ionic equation, and then calcium chloride breaks up into calcium ion and chloride ion. Now, remember, we're gonna swap Ionic Partners because opposite charges attract this. Plus one charge is attracted to this minus one charge. So when they connect together, remember when the numbers are the same, they just simply cancel out. So here, that will give me an H four c l plus, Then we're gonna have here calcium ion and sulfate ion. The numbers and the charges again are the same. So they're going to cancel out. Give me calcium sulfate. At this point, we have to remember our saw You bility rules to see if we created Ah, solid liquid or gas. Otherwise, no reaction has occurred. So remember, based on our side ability rules, anything connected to the ammonium ion will be soluble. So this is a quiz, and then we have calcium sulfate based on the rules that we learned about calcium sulfate, remember, it's gonna form a precipitate if sulfate ion is connected to CBS. Remember, CBS stands for calcium, barium or strontium. Since sulfate here is connected to calcium, it's gonna form a precipitate. Since we made a solid, a reaction has occurred. Next we just have to balance out this molecular equation. We have to. Ammonium is here and to chorines here, So I have to put a two right here. The sulfate There's just one sulfate and one sulfate, one calcium and one calcium. So here we have coefficients of one for the other compounds. Now we're going to go to step five where we break up on Lee the acquis compounds into the respective ions. So everyone breaks up into ions except for calcium sulfate, which is a solid. So here remember, the coefficient gets distributed to each of these ions we're gonna have here one ammonium ion, acquis. But remember this little to here, so there's actually two ammonium ions. Plus, we're gonna have sulfate ion acquis plus one calcium ion. Acquis. Plus, we have a two here to chloride ions. A quiz gives us The two gets distributed to each one of these ions here, so we're gonna have to ammonium ions. Acquis plus two chloride ions. A quiz plus one. Calcium sulfate solid. Here. We're gonna cancel out the spectator ions from the complete Ionic equation in orderto isolate the Net Ionic equation. At this point, what we have is the complete ionic equation for the net. Ionic, Let's remove the spectator ions. Remember, spectator ions are the compounds that exists is both reactant and products at the same time. So here, ammonium ion here matches up with ammonium ion here. So those will not be part of the net. Ionic Also we have here to chlorides and to chlorides. They also will not be part of the net Ionic. That means all that's left at the end. When we bring things down, will be s 04 to minus acquis, plus calcium. Ion acquis gives me calcium sulfate solid. So this year represents my Net Ionic equation. So just remember a lot of what we learned in terms of obtaining the Net Ionic equation has the first deal with determining the molecular equation. From there, we have to look at the complete Ionic equation and by removing the spectator ions, that's how we're able to isolate our Net Ionic equation at the very end.