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36. Write the complete molccular; total ionic and netionic equatians for the reaction of Iithium carbonate and cobalt (II) nitrate. Stotes 01 matter QEst be induded...

Question

36. Write the complete molccular; total ionic and netionic equatians for the reaction of Iithium carbonate and cobalt (II) nitrate. Stotes 01 matter QEst be induded in all equations The composition of the precipitate must be predicted. Use the so ubxlity cnart provided In the Exam notification Or your reachons ab: (6)Complete equation;Total lonic cquation: Net lonic equation:HTMLEditor1 " 4*4-I00020 * * 00 M - D ^ % Tent Paragraph0 KcrdlaMMacBockAur883

36. Write the complete molccular; total ionic and netionic equatians for the reaction of Iithium carbonate and cobalt (II) nitrate. Stotes 01 matter QEst be induded in all equations The composition of the precipitate must be predicted. Use the so ubxlity cnart provided In the Exam notification Or your reachons ab: (6) Complete equation; Total lonic cquation: Net lonic equation: HTMLEditor 1 " 4*4-I00020 * * 00 M - D ^ % Tent Paragraph 0 Kcrdla MMacBockAur 883



Answers

Write balanced complete ionic and net ionic equations for each reaction. (a) $\mathrm{AgNO}_{3}(a q)+\mathrm{KCl}(a q) \rightarrow \mathrm{AgCl}(s)+\mathrm{KNO}_{3}(a q)$ (b) $\operatorname{CaS}(a q)+\operatorname{CuCl}_{2}(a q) \longrightarrow \operatorname{CuS}(s)+\operatorname{CaCl}_{2}(a q)$ (c) $\mathrm{NaOH}(a q)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaNO}_{3}(a q)$ (d) $2 \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{NiCl}_{2}(a q) \longrightarrow$ $\mathrm{Ni}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{KCl}(a q)$

So for the for the given question the complete ionic equation will be to potash um positive plus C. 03 to negative plus. See you too positive plus two N. +03 negative leads to formation of ceo. See you see oh three less to petition positive plus two N. 03 negative. Whereas the net my neck equation will be See You too positive plus ceo tree to negative leads to formation of see you C. 03 See that the sum of charges on both the sides of net equation is equal. We have Plus two and -2 Which leads to overall charge on the product side as zero. And the name of the precipitate formed will be corporal two Carboni.

From the following reactions. Let's right the Net Ionic equation for any precipitates that air formed for a We see that we have a precipitate of iron carbonates so or Net Ionic equation would be f e three plus at three to minus she producer precipitate of iron carbonates. It's on balance. It for be, will produce a precipitate of mercury chloride h g two she plus bad CR minus to produce our precipitate of h g to C L two solid to there to balance perceive will produce a producer precipitate of copper. Oh, sorry. Um, for C, there is no reaction As both products are soluble. There's no precipitate for d will produce a precipitate of copper sulfide precipitated copper So fine for e Well, pretty soon precipitate of lead chloride. Great f producer precipitated calcium carbonates and for G We're pretty, sir. Precipitated gold hydroxide. So here are all of the net ionic equations for the precipitates that form from the reacting given

So let's take a look at the reaction between calcium hydroxide and cobalt two chloride. That's going to form calcium chloride and cobalt two hydroxide. First thing we do when we want to determine complete and net ionic equations is to first balance this molecular equation in this case it is already balanced. Next, we're gonna want to check our scalability rules and see whether these substances uh are soluble or insoluble. And so calcium hydroxide when you look at it as insoluble. So that will be a solid Cobalt two chloride is soluble or acquis, it will dissolve into its ions. And calcium chloride will also dissolve and break apart into its ions. And Cobalt two hydroxide does not. It is a solid. So this is important because now we know that the cobalt chloride and the calcium chloride will break apart into ions. So let's go ahead and make the complete ionic equation for this solids stay together. So we will have our calcium hydroxide. The ions here break apart. So we're going to have cobalt As a two plus charge. And we're going to have to chloride ions. And both of these are acquis, they've dissolved. And here we're going to create the calcium ion. Two chloride ions, both acquis And the solid cobalt two hydroxide will state again, this is the complete ionic equation. Now to make the net ionic equation we remove the spectator ions when we do that. And so spectator ions are those that are not participating in the reaction and we know they're not participating if they don't change from the reactant side to the product side. And so you can see these two chloride ions are the same on both sides. They haven't changed. So it's like subtracting two chlorides from both sides. Those are our Spectators. And now the net ionic equation is everything that's left. And so we have our calcium hydroxide which is a solid cobalt ion Aquarius, our two chloride ions a quick I'm sorry those are not part of it. We're going to go to calcium ion equals And Our Cobalt two Hydroxy. So this is our net ionic equation. So let's take a look at another example. Let's say that we have barium chloride and we have sodium carbonate and that goes to make barium carbonate and sodium chloride. The first thing we need to do is look to balance the equation and we need to sodium is and to chlorine is on the right. So the two that balances our equation. Next we have to check our sally ability rules. And so when we look up barium chloride and sodium carbonate we find that both of those are soluble, they're acquis they dissolve into their ions in water. And then when we take a look at the product side you'll find that barium carbonate is insoluble. So it's a solid but sodium chloride does dissolve. It's a quiz. So this is our molecular equation. And now we're going to go ahead and find our complete ionic equation. So we have to go find the charges of these substances and know the number that we have. So barium will be an ion two. Plus we have to see L's chloride is C L minus. We have to sodium seems and we have one carbonate Which is -2. The barium carbonate stays together and the sodium chloride breaks apart. But now if you notice here there's a two coefficient. So that means that there's two sodium and two chlorides. So this is our complete ionic equation. If we want to go and find the net ionic equation, we are going to have to remove the spectator ions. Spectator ions do not participate in the reaction. So you can see that we have to chlorides on both sides. So we will go ahead and remove both chlorides. You can see we have two sodium is on both sides. So we will go ahead and remove the two sodium. And so the net ionic equation is whatever is left. We have our barium ion Aquarius and our carbonate ion Equus and when they come together they are going to form solid barium carbonate. So that is our net ionic equation. For a second example, let's go ahead and try one more example. So let's go ahead and take sodium phosphate and to that we will react with nickel to nitrate. That's going to make nickel to phosphate and sodium nitrate. So this is our molecular equation unbalanced. We have to go ahead and balance the equation and we can balance the equation with the coefficients to 31 six trump. Now, once we have that we can go ahead and see which substances are soluble and which ones are not. And it turns out here that we have our sodium phosphate. Based on our rules is soluble or Aquarius. R Nicol to nitrate is soluble or acquis R Nicol to phosphate is insoluble or a solid. And our sodium nitrate is a quiz. So we can go ahead and come up with our complete ionic equation. All right, So if you take a look here, we have two sodium phosphates. Each sodium phosphate has three sodium. So there's two times three or six sodium ions which are Aquarius. We have two times one phosphate or two phosphates. The 3- Charge that our acquis We have three nickels with a plus to charge Equus and we have three times two or six nitrates which are acquis. We do the reaction we're gonna form are nickel too fast feet, we're going to have six sodium size Equus And we're gonna have six nitrates that are eight ways. And that will wind up being are complete ionic equation. Then we have to go and figure out which substances are Spectators and remove those to make our net ionic equation. So there are six podiums on the left, there are six on the right, so they will cancel out. There are six nitrates on the right, there are six on the left. They will cancel out and we're left. Then with our net ionic equation, I'm just gonna put the positive 1 1st because that's how we Make compounds in the end. So I'll take our three Nichols that our nucleus With our two plus charge our two phosphates three minus charged. And we'll create our nickel to phosphate solid. Let me just make sure I put a solid on the complete one as well. All right, so here are three examples of making complete and net ionic equations from our molecular equation.

In this problem, we are going to write some net ionic equations and to determine if a reaction occurs based on solid ability rules. So first we're looking at this reaction with sodium. I owns pro my lions led irons and nitrate ions in Solution. So looking at our scalability rules, sodium and nitrates are generally soluble, so these will not precipitate. However, lead is generally insoluble, and with bromide ions, they will precipitate together. So are net. Ionic equation will look something like this. Um, these four. Together we'll have our sodium ions that remain in solution. We'll have our nitrate ions that remain in solution, and then we will have our lead bromide precipitate. Since we had a two here, let's just balance this and assuming that it was a sodium bromide next year, this will be our net ionic reaction, and this is our precipitate. For the second solution. We have a magnesium chloride solution and copper sulfate solution mixed together, um, trying to figure out which ones will precipitate. Looking at our scalability rules. Coloreds air. Generally soluble except with heavy metals, sulfates are also generally soluble, except with some group or heavy group. Two elements magnesium is group to, but it is a light group to element, So none of these election precipitated. There will be no reaction here, so no reaction. The Net Ionic equation will just be the same thing. Is this right? All of these will exist, uh, as ions in solution for part C, we have iron nitrate mixed with sodium hydroxide. So again, sodium is generally soluble a group one element. Nitrates are also generally soluble. Hadrt oxides, however, are generally insoluble, and iron will precipitate with the hydroxide. So are net ionic equation. We will have our so diem's that remain in solution. We'll have our nitrates that remain in solution and we will have our I own hydroxide that precipitates from the solution, and that is our net ionic equation.


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