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Question

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Answers

For the following chemical reactions, determine the precipitate produced when the two reactants listed below are mixed together. Indicate “none” if no precipitate will form
$\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{3} \mathrm{PO}_{4}(a q) \longrightarrow$__________________(s)
$\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow$__________________(s)
$\mathrm{NaCl}(a q)+\mathrm{KNO}_{3}(a q) \longrightarrow$__________________(s)
$\mathrm{KCl}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow$__________________(s)
$\mathrm{FeCl}_{3}(a q)+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow$__________________(s)

For our first example, we're going to take nickel to nitrate, which is an agreed solution and add it to an equally solution of NH so our two products are sodium nitrate and nickel to hydroxide, which if you look at your scalability rules, you can see that that is insoluble. So there's our balanced equation. And this is our precipitate. Right here. In our second example, we're going to take our acquis sodium hydroxide and reacted with Aquarius potassium sulfate. Okay, this is going to give no precipitate. Alkaline metals are genuinely soluble so we won't get a precipitate from that. So we're going to take an Aquarius solution of sodium sulfide and add it to our copper to acetate. Yeah. Mhm. So we're going to form some sodium a city. Okay? It has an alkali metal in it, so that's Aquarius. And then we're going to make some copper to sulfide, which based on our side liability rules is definitely soluble. We'll go ahead and balance this and this is our precipitate.

To obtain a net ionic equation. It's helpful to first write the complete Molecular or complete formula equation. one balance it, then separate everything into ions. To get the complete ionic equation, then cancel the spectator ions and what is left over will be the net ionic equation. So for this question, you first need to identify if a reaction even occurs By using the cell ability rules found in Section 43. A reaction will only occur if one of the potential products is insoluble. To identify the potential products, you simply switch the cat eye on places. So if we have calcium chloride reacting with potassium phosphate, one potential product will be potassium going with chloride and then calcium going with phosphate, calcium has the two plus charge, phosphate has a three minus charge. So to get a neutral compound, we're going to need three calcium and to phosphates according to the soluble itty rules found in section 43 This is insoluble so we'll write it as a solid. The next product would be potassium react or combining with chloride. potassium has a plus one charge. Chloride has a one minus charge. So we just need one of each of them. But now we need to balance the chemical reaction. We have three potassium and one potassium. We have to calcium, one, sorry, 3 calcium and one calcium. So let's start by putting a three on the calcium. We have to phosphates and one phosphate. So let's now put a two in front of the potassium phosphate. That gives us six potassium and six chlorides. So we now need to put a six there. The next step is to separate the ions from the ionic compounds to create the ionic equation, the complete ionic equation, calcium chloride, three moles of it will separate into three most calcium and six months chloride, potassium phosphate, two moles of it will separate into six moles, potassium and two moles phosphate. The calcium phosphates is solid so it stays together and the six moles potassium chloride will separate into six molds, potassium and six months chloride. We then see that potassium six moles and chloride. Six moles are comment to both sides so we'll cancel these ions. These are the spectator ions. And what is left is now our net ionic equation three moles calcium two plus react with two moles phosphate, three minus to produce calcium phosphate. Now, for part B we have potassium chloride reacting with sodium sulfate. If the cat eye on switch places than sodium will combine with chloride, sodium has a plus one charge, chloride has a one minus charge. So we just need one of each of them in the correct chemical formula. According to sally ability rules this is soluble so it will stay as Aquarius. Then potassium can combine with sulfate. potassium has a plus one charge. Sulfate has a tu minus charge. So we need to potassium for every sulfate and according to sally ability rules both of these are soluble because they are both soluble, they are both Aquarius and no reaction occurs because no precipitate occurs to balance it. We simply need to put it to in front of the sodium chloride and potassium chloride. But there is no net ionic equation for part B. Because there is no precipitate that is formed for part C. We have ammonium carbonate reacting with barium nitrate. And when the Catalan switch places, barium carbonate is a potential product and according to the scalability rules this is insoluble. So it will be a solid. This will be the precipitate that forms. And then ammonium nitrate. Ammonium has the plus one charge. Nitrate has one minus charge. We just need one of each of them. But then to balance it we've got to ammonium is to nitrates So we're gonna need to put it to their one barium one carbonate. So it is now balanced. When ammonium carbonate separates. We get to ammonium is one carbonate. When barium nitrate separates we get one barium to nitrates. Barium carbonate stays together and then we get to ammonium zand to nitrates. When two ammonium nitrate separate, the nitrates are common to both sides. They're spectator ions that cancel. And what is left is barium two plus reacting with carbonate to minus producing barium carbonate solid as our precipitate. Mhm. For d. We have iron to chloride reacting with potassium hydroxide. When the Catalan switch places we will get iron to hydroxide and according to the scalability rules hydroxide. Czar insoluble. Unless there with potassium nitrate er sorry potassium sodium or lithium which iron is not. So this is our precipitate And then potassium will go with chloride, potassium with a one plus. Charge chloride with a 1- requires one of each of them. And then to balance it we have to chlorides so we're going to need to put it to here giving us to potassium so we need to put it to here giving us to hydroxide but we already have two hydroxy sides. So then we'll separate the iron to chloride into iron two plus and two chlorides. And then to potassium hydroxide separate into two potassium ions and two hydroxide and ions. Iron to hydroxide stays together as the precipitate and then to potassium chloride separate into two potassium ions and two chloride and ions. The potassium ions are common to both sites. There spectator ions. Chloride ions are commendable sites. Also spectator ions. So what is left then is the net ionic equation. V two plus reacts with two. Hydroxide is producing iron to hydroxide solid. Yeah, for e we have barium nitrate reacting with sodium hydroxide when the Catalan switch places bury him. We'll go with hydroxide. Barium has the two plus charge. Hydroxide has a one minus. So we're going to need to hydra oxides and according to Saudi ability rules this is a hydroxide so it's insoluble unless it's a sodium lithium or potassium hydroxide and then sodium with a one plus charge, combines with nitrate with a one minus charge. So we just need one of each of them. And according to the scalability rules, anything with sodium nitrate this contains both is soluble. To balance it, we have to nitrates. So we needed to put it to hear that gave us to sodium so we needed to put it to hear that gave us to hydroxide. We have two hydroxide. one barium. It's now balanced. This is then are precipitated barium hydroxide. When barium nitrate separates, we have one barium to nitrates. When two sodium hydroxide separate, we have to sodium two hydroxide barium hydroxide stays together. And then when two sodium nitrate separate, we get to sodium and to nitrates the sodium Z and the nitrates. Two of them are common to both sides. These are the spectator ions. And then what is left after removing the spectator ions from the ionic equation is the net ionic equation. One barium two plus reacts with two. Hydroxide is going to bury him. Hydroxide solid. Okay now for f. We have sodium sulfide reacting with antimony. Three chloride. When the cat eye on switch places we'll get antimony sulfide. Antimony has a three plus charge. We know that because Gloria has a one minus, there are three of them. So antimony must have a three plus. Sulfide has a tu minus. We know that to sodium has a one plus there are two of them. So sulfide needs to be to minus. So if we have a three plus, combining with the tu minus, we're going to need two of the three pluses and three of the two minuses. So antimony sulfide will take on this chemical formula and according to the scalability rules, sulfides are, most sulfides are insoluble and this is not one of the exceptions. And then sodium goes with chloride, sodium with a one plus charge chloride with 1- requires just one of each of them. So don't get in the habit of saying N A to C L. Three because that's not a compound, sodium chloride is N A C L. We now balance it to take care of the multiple chlorides and sodium we have to antimony so we'll put it to their We have three sulfides. So we'll put a three there. That then gives us six sodium and six chlorides. So we need six sodium chlorides. Our precipitate. Then his antimony claure or antimony sulfide. When three sodium sulfide separate we get six sodium and three sulfides. When to antimony chloride separate we get to antimony knees and six chlorides. Antimony sulfide stays together. It's our precipitate. And then when six sodium chloride separate we get six sodium says and six chlorides. These sodium and chlorides are our spectator ions, comment to both sides. We cancel them. And what is left is our net ionic equation to antimony three plus is combined with three sulfide two minuses to give us antimony sulfide precipitate. Now for the last one we have led to nitrate reacting with potassium sulfate. When the Catalan switch places we will get led to sulphate. Lead with a two plus sulfate with a tu minus gives us just one of each of them. Most sulfates are soluble but led to sulfate is an exception. So this is insoluble, this is our precipitate and then we get potassium with a one plus charge, combining with nitrate with a one minus charge. So we just get one of each of them in the chemical formula. How to balance the chemical reaction we have two potassium. So we had one so we put a two there that gave us two potassium introducing to nitrates but we have to nitrates, one LED one led one sulfate, one sulfate so it's balanced. We'll separate the lead to nitrate into one lead two plus and to nitrates and then potassium sulfate is separated into two potassium and one sulfate led to sulphate stays together as a solid and then the two potassium nitrate separate into two potassium is and to nitrates. The potassium and the nitrates are common to both sides, so they cancel when they council what is left our our spectator ions. I'm sorry these are our spectator ions. What is left is the net ionic equation Lead two plus, reacting with sulfate, producing lead to sulphate solid.

So in this video, we're gonna go over Question 52 from Chapter four, which size when the following solutions are mixed together. What precipitate if any will form. So when a we have ht to ino 32 plus C u S 04 Um, So what products could we possibly form? Well, we could form copper to nitrate or mercury one sulfate those air the two products we could form. And since most nitrate salts are soluble, coppers to nitrate is soluble and most soul fights assaults air soluble. But the exception is mercury one sulfate. So this is insoluble. So we are going to form a precipitate, and it will be the mercury one sulfate and B. We're given nickel to nitrate plus calcium chloride. So, um, what can we what products can we possibly form? Will of our nickel can tie on, combines with their chloride, and I on we get nickel to chloride. And if our calcium cat eye on combines with their nitrate and I on, we get called on calcium nitrate. So since most of our chloride salts are soluble, um, then our nickel to chloride is valuable. And since most of our nitrate salts, air soluble are calcium nitrate is also soluble, So we're not gonna form any precipitate and B and see we're given Que tu sio three plus m g. I too. So potassium carbonate plus magnesium iodide. So what products can reform will live far Potassium cat eye on combines with our iodide and I only get potassium iodide. And since most of our iodide salt are soluble than this is valuable, um And then if we combine our magnesium cat eye on with our carbonate and I on, we get magnesium carbonate and most of our carbonate salts are only slightly soluble, except for those with certain cat ions. But my museum is not one of those cat eye on, so this is only slightly soluble, so we'll form some purchase, um, precipitate and that precipitate will be magnesium carbonate in d were given, um, sodium chrome eight plus aluminum bromide. So what products can we form? Well, if our sodium cat eye on combined with their bromide, and I only get sodium bromine And since most of our bromide salts are soluble, this is soluble, um And then if we combine our aluminum cat eye on with our crow, my and I on. Then we get aluminum crow, mate. Um, and most of our chrome eight salts are only slightly soluble except where those with certain cat I owns. But aluminum is not one of those cat eye on, so this is only slightly soluble, so we'll form some precipitate on that precipitate will be aluminum crow, mate.

Chapter six Problem 48 gives us a variety of solutions mixed together and asks which will form precipitate so to solve these. First, we'll look at the different products that could form and then see if any of them are solid or it's in soluble in water. Those that are insoluble will be the ones that form a precipitate. So let's start with this first problem in order to see what the potential products could be, we will exchange the ions here. Essentially H g will switch with See you. We do that. We find that we have HD s 04 and see you and no three to now we need to determine if either of these is solid or insoluble in water. If we look at this eligibility table, we find that H G s 04 is in fact solid. Therefore, we do have a precipitate forming specifically the precipitate HD as so four solid. Now let's look at the next problem here we have and I know 32 and c a c l two. So again it will switch our ions here. And that will give us and I, c e o or C A and No. Three. If we look at a soluble ity chart, however, we see that neither of the's is a solid precipitate were insoluble in water. Therefore, for this problem, we have no precipitate. Now let's move on to see again we'll switch our ions So we have MGI c 03 plus k I. And if we look at a soluble ity table, we find that MG CEO three doesn't fact form a solid precipitate. So we have a precipitate of m G c 03 Finally, let's look at the last one again. We will switch our ions giving us a l c R O for three with a little too. And again To get those numbers, we will have to be able to balance out all of our charges for each molecule. And that also gives us and a BR once again looking at our soy ability table, we find that ailed to C R O for three does form a solid for


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