Lecture 25. Review Final Part II
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River sourceAzo couplingMoleculeAtomic numberSet (abstract data type)Lecture/Conference
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Growth mediumComputer animation
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Man pageIce frontGolgi apparatusHydroxybuttersäure <gamma->Electronic cigaretteCobaltoxideAgeingChemical elementSea levelThermoformingChemical structureChemical formulaCondensationPlant breedingFunctional groupHydroxylKohlenstoff-14ChemistryMaßanalyseAtomic numberCarbon (fiber)CarbonHydrogenWursthülleChain (unit)MoleculeAttachment theoryLecture/ConferenceComputer animation
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SulfateSolutionReaktionsgleichungExplosionStoichiometryPrecipitation (chemistry)SolventFunctional groupHuman body temperatureVerwitterungBacterial artificial chromosomeSolutionThermoformingKaliumsulfatSong of SongsPeriodateBariumReaktionsgleichungData conversionPotassiumSemioticsMolar volumeAqueous solutionAmount of substanceCatalytic converterChlorineBarium sulfateBarium chlorideKaliumchloridChemical formulaSulfateMoleculeSulfateLecture/ConferenceComputer animation
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ReaktionsgleichungSulfateSolutionMan pageSodiumAcidPotassiumWursthüllePrecipitation (chemistry)Molar volumeBase (chemistry)SodiumChemical reactionMolekulargewichtsbestimmungBarium sulfateSolutionKaliumsulfatSodium hydroxideHydroxideElectronic cigaretteWaterFormaldehydeWearFrench friesLecture/ConferenceComputer animation
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SolutionAcidReaktionsgleichungWhitewaterAcidChemical reactionHydroxideBase (chemistry)AgeingSolutionRepeated sequence (DNA)ThermoformingNeotenyWaterSteelPerchlorateAmount of substanceSodium hydroxideSodiumChemical compoundLecture/ConferenceComputer animation
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SolutionAcidData conversionCatalytic converterWalkingMolar volumeWine tasting descriptorsLecture/ConferenceComputer animation
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AcidSolutionSodiumSulfurGasReaktionsgleichungMolar volumePerchloric acidHydroxideWaterfallWalkingRiver sourceChemical reactorRedoxSense DistrictLecture/ConferenceComputer animation
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SulfurGasReaktionsgleichungSolutionMortality rateRedoxProtonationHydrogenElectronCHARGE syndromeOxideAgeingStickstoffatomSolutionJadeMan pageElectronic cigaretteElimination reactionReducing agentSulfurChemical elementOxideWalkingMoleculeCobaltoxideNitrateSpectroscopyAtomic numberReaktionsgleichungAzo couplingSulfur dioxideStickstoffmonoxidSodium nitrateThermoformingStoichiometrySodiumChemical compoundGasWaterIce frontLecture/ConferenceComputer animation
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RedoxSulfurSolutionGasReaktionsgleichungBerylliumMan pageSense DistrictLactitolExplosionSolutionRiver sourceWursthülleCHARGE syndromeUmweltchemikalieElectronic cigaretteActive siteCalciumBrominePrecipitation (chemistry)IonenbindungSodiumMoleculeSulfurElectronWaterStickstoffatomCalcium bromideSulfateOxideStickstoffmonoxidBerylliumChemical reactionThermoformingHydrogenReaktionsgleichungCobaltoxideCalcium sulfateMolar volumeAcidLecture/ConferenceComputer animation
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BerylliumElutionSolutionMan pageFormaldehydeVolumetric flow rateReaktionsgleichungMagnesiumInclusion (mineral)Chemical compoundCHARGE syndromeSolutionSea levelAgeingReaktionsgleichungWine tasting descriptorsOceanNeutralization (chemistry)Chemical reactionExplosionFiningsWhitewaterThermoformingMedical historyWaterfallCell growthGlassesLightningCalcium sulfateMolekulargewichtsbestimmungMagnesium hydroxideMagnesiumWalkingBromwasserstoffsäureAbbruchreaktionPrecipitation (chemistry)Azo couplingBerylliumStuffingBariumCalciumCalculus (medicine)Amount of substanceWursthülleGrading (tumors)Molar volumeLevomethadonBeryllium sulfateCalcium bromideHydroxideWaterAqueous solutionChemical experimentLecture/ConferenceComputer animation
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Inclusion (mineral)SolutionMagnesiumRedoxErdölraffinationThermoformingElectronElectronic cigaretteWaterfallBranntweinZincAgeingExsudatSaltCopperSolutionOxideWalkingSulfateAtomChemical elementZinc sulfateCopper(II) sulfateMagnesium hydroxideMagnesiumStoichiometryStuffingMolar volumeHydroxideCobaltoxideSulfurNeutralization (chemistry)WursthülleReducing agentAmount of substanceRedoxComputer animation
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MashingSolutionAtomRedoxLecture/Conference
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Ballistic traumaComputer animation
Transcript: English(auto-generated)
00:06
All right. Friends. All good things come to an end. It's true. It's true. And we started out almost three months ago
00:24
as strangers now we're all friends, right? Wonderful. And in the meantime, I actually learned a couple of things about atoms and molecules. And so we'll just try to do it the best we can next time. Friday, when we will meet once again in a slightly different setting. You guys
00:45
will be doing the exam and I'll be facing around and grading the exams afterwards. Okay. So this lecture just reviewed some additional points that we didn't get to last time. So over some of
01:02
the things, I had a couple of issues. And then hopefully there will be some kind of practice. Only other practice. I think should be addressed, and will be addressed, during Monday's review session. And this picture shows a person who studied too hard, okay, so studying is pretty important. We'll make sure you are
01:32
in good shape, but I don't want to do it. I have folks that think that it would be important to do an exam I have to study through the entire night, and don't sleep. And I don't think that's a good way to do it.
01:47
Because this is what it will look like, and this is what the exam will look like, too. You need your brain your brain needs rest to function well. So
02:01
make sure to attend the Monday previous session at 7 o'clock, and it's the final service that we offer next week, because it's the final week, the very odd week, it's not a regular week, so there will be no office hours, especially if it's a office hour sometimes. They're only there. So this is the last kind of opportunity to interact with your teaching team
02:26
beyond the form, which will be open 24-7. So we will keep conversing there. Okay. So let me go back a little bit and talk about
02:42
these kind of questions where we look at molecules, and we look at ways in which to write down the formula, molecular formulas of molecules like this. This is a line structure and we should be able to draw line structures, not as complicated as this, but we should be able to write down the molecular formula from this.
03:05
Okay? The chemical formula. So the molecular formula, the chemical formula, is something that says C, X, H, Y, and so forth. So we have to count the number of atoms of each kind, and then write that down in a molecular formula. That is what you have to do. So it's quite simple.
03:26
Let's first look at the C's. There are 8. Why? Because there's one here, one there. one there, one there, one here, one there, one there. And one right there. This end point is a carbon atom with three hydrogens attached. Remember?
03:43
So that's a total of eight. And the next element we are to put down is the H. The hydrogen. there's one here, one there, one there, two here, two there, none here, none there, none, one here, and there's three there. That's the total of a level.
04:05
Then there's two more elements, in this case there's one, two, three, oxygen and there is one, nitrogen. And then you're done. What should you not do? Well, you shouldn't start writing here HOC
04:21
H2 C, and I'm just like having this long list of elements in there. No. This is a molecular formula. We're not asking you to invent your own fantasy condensed structure or formula of a structure like this you will not succeed, and you will see an ugly red cross in your answer on the exam.
04:43
Okay? The molecular formula looks like this. It's compact, it only lists the elements with the count of each element. That is the answer. Another kind of question is I list a condensed structural formula. This is a condensed structural formula. This is not
05:03
called the molecular chemical formula. This is called the condensed structural formula. Okay? This, we should be able to translate into an actual line structure. So let's do that. First let's have a look at this thing
05:21
when we start to read it. From the left to the right, I see two CH3 groups. The CH3 group looks like a stick. These two sticks are connected to this carbon atom, which also happens to have another H, which I will not draw, because H is another H. So this carbon atom here has
05:42
one CH3 group attached, another CH3 group attached, that's why there's two here, one H, and one carbon next to it. This carbon has an oxygen, and another neighbor, which is carbon. This carbon has H2 and no H. So it looks like this.
06:01
Okay? This point here is that carbon right there. It has one, two lines, those are the two CH3 groups. This CO is this one here. C with one oxygen attached. On the other side, it has one carbon, and another carbon. This is the carbon with
06:23
oxygen, on one side it has this carbon, on the other side it has that carbon right there. This last carbon, all the way to the right is this carbon, the last carbon in the chain. and that's one H, sorry, that's two H's, one, two, and an OH group. Which is this guy right here.
06:44
So this is very systematic. It's very systematic. So make sure you are able to translate this into a formula like this. If you've done many of these, and you can go back to the midterm, to the first midterm, remember there's four midterms there with answers for you to look at. So it's a good practicing term.
07:10
Okay. Let's now move on to precipitation reactions in particular. Let's look at the stoichiometry of these things. So a typical question could go like this
07:22
one mole of an aqueous potassium sulfate solution reacts with four moles of aqueous barium chloride solution and then I have to calculate the amount of grams of the precipitate typical, typical question. two solutions are given, okay, the molarities are not given. The volumes are not given, the molarities are not
07:43
given, but it is different how many moles I have. That's great. I already have moles. That's what I want. I have more moles, because moles allow me to make mole ratio conversions. Okay? conversion through mole ratio. I have to calculate the amount of precipitate formed. So I have to calculate how many moles of precipitate do I form, then convert moles into grams.
08:08
So first things first, I want to write down a balance equation. I just have to translate this word into a chemical formula. Potassium sulfate. Potassium is K, sulfate is SO4 2-, and it is K2 SO4 2. Sorry, SO4.
08:29
That's right. This one here, barium chloride, barium is 2+, chlorine is minus 1, so it's BaCl2. Okay? So here it is, K2SO4 that is potassium sulfate
08:43
this is BaCl2. That is barium chloride. You can make two new combinations, okay, namely, potassium with the chlorine ion, that will not form a precipitate. Okay? Because chlorine dissolves well with many cations, and in particular with cations that are of polymetals.
09:07
Okay? So your polymetal ions dissolve with almost everything. So this guy will not form a precipitate. Therefore, potassium chloride is fully dissolved in the solution, therefore the only possibility for a precipitate is barium sulfate.
09:23
And indeed, sulfates typically dissolve well, but not with barium. Barium wasn't acceptable. On your table, you can read that. So barium sulfate is a precipitate that gets an S, while this has an aqueous symbol. Okay? So this is the equation. It's actually the molecular equation.
09:44
Given in terms of molecules. Then I'd like to determine the limiting reagent. So I already have the number of moles and so I'm just going to do a quick comparison with the balanced mole ratio versus the actual mole ratio. So the equation says this.
10:00
One mole of potassium sulfate for each one mole of barium chloride. That's what I mean. Okay? That is one to one. Let's look at the actual numbers. This is easy, because the number of moles is given right there. If I divide one by the other, I put, of course, potassium sulfate on top. How many? One mole of potassium sulfate,
10:22
four moles of barium chloride, so it looks like this. And that is less than one, and therefore, the numerator is the limiting reagent. K2SO4 is the limiting reagent. How long does it take to determine the limiting reagent?
10:44
It doesn't have to take very long. Because this is what you have to do. One ratio, another ratio, a lukewarm which is bigger, boom, you know which one is the limiting reagent. Don't make that, you know, a 30-minute search on your exam. This literally takes one or two minutes to figure it out.
11:04
Once you know the limiting reagent, the limiting reagent, of course, determines how much precipitate you form. It determines the amount of the products. Because what I'm going to do is I'm going to take the number of moles, in this case one, of the limiting reagent, which is K2SO4,
11:21
one mole of K2SO4, I'm going to convert into moles of the product. The product is barium sulfate. The ratio between barium sulfate and potassium sulfate is one to one. That's what the equation says. With each one mole of potassium sulfate, you make one mole of barium sulfate. And therefore,
11:42
one to one, and now I convert this one mole of potassium sulfate into moles of barium sulfate. Once I know the moles of the product, I simply multiply that by the molar mass to find grams. And here's the molar mass of barium sulfate.
12:01
233.43 So you can see, if you look at the units of molar mass, it's grams per mole. This mole very nicely crosses out. So you convert your answer into grams of barium sulfate. The answer is 233 grams of barium sulfate. Okay. Let's do another one. Let's do an acid-base reaction.
12:35
Okay. The question is a typical question. And I think something similar actually appeared on the midterm.
12:42
How many milliliters does a volume, so what is the volume of a particular solution that I have to add to an existing basic solution, sodium hydroxide, to neutralize this solution? So let's visualize it. Let's visualize the situation. If you read a question like this, you want to make
13:03
sure you understand the physical situation. I have a bucket with, what is it? Sodium hydroxide. It's a basic solution. The basic solution has volume and molarity. Is that given? Yes. That's not exactly given, but it's a basic solution. It's given how many grams there are. So I know how many OH- ions there are in the solution. The question is, how much of
13:25
an acid do you have to pour in there, what's the volume of the acid you have to pour in that bucket to neutralize all the OH- ions? That's the question. Okay? The acid that you are going to pour is perchloric acid. So a reaction is going to take place.
13:50
Okay. The first thing we'd like to do is to balance this equation. This is sodium hydroxide.
14:03
Oh, interestingly, it is solid. So in a bucket, it's not even water. It's just a chunk of sodium hydroxide. It's solid. It has an acid on top to neutralize that. It has an S here. The fact that this is not in solution
14:20
makes no difference for the reaction that's going to take place. Both of these are strong bases and acids. So this is a strong base, this is a strong acid, this H plus will attack the OH- whether it's in a solid or in a solution, it doesn't matter.
14:42
So this H plus will attack this OH- to form water. And then the other ions are ClO4- which is the perchlorate anion, and the sodium plus tetrion. These two things remain heavily in solution, both dissolve very well.
15:00
So this is your rest ions if you want. So the actual reaction simply is this H jumps on the OH and forms water. Looking at this, you can see immediately that for each OH in my bucket, I need one H plus. So what I'm going to do is this.
15:25
I say I need to neutralize this OH- part of this solid. So in order to do that, I need to know how many of the OH-s are there. Do I know that? Yes, I know. Because I have here a mass and I can convert mass into moles. So let's do that.
15:42
2.5 grams this is the mass of sodium hydroxide and I get 0.02 sorry, 0.0625 moles of sodium hydroxide. That's how many moles of sodium hydroxide units I have. What I want is to know how many OH-s I have.
16:03
Okay? But that's the same. Because each sodium hydroxide has only one hydroxide, that means you have the same amount of hydroxide. Okay, so this is my target, I have to get rid of this by adding H pluses.
16:21
So how much do we need? Well, I have this many they react one to one okay, so for each OH- I need one H+, so this OH- is crossing out and I convert into the moles of H+. and each HCl4 has one H- to offer.
16:43
So I can convert from H- into this compound here, the chloric acid. Each one mole of chloric acid has one mole of H+. So what I do, I convert from OH- to H+, to HCl4. Okay? So this is basically just one, one, one that doesn't react all the same.
17:04
I need this many moles this many moles of chloric acid. I calculate the number of moles of OH-, I can calculate the number of H+, and how much HCl4 I need. HCl4. So this is the number of moles, that's not the answer yet, because I could
17:21
tell the person that's ready my exam how many milliliters I need. So I have to convert the number of moles into volume. moles to volume. Is that a conversion you can make? Yes. You can, if you know molarity. Is molarity given? Yes. It's given. That means I can do one step to convert moles into volume.
17:46
Okay? So let's do that. This is moles that's converted to volume. I'm going to use this. So divide that molarity. Why? Let's look at the units. Molarity is mole over liters, but I have to divide the molarity so that the mole appears at the bottom, and the liter appears on top, because I want to go to volume.
18:05
Okay? So this number divided by molarity gives me something in terms of liters I want milliliters, so I do a last conversion here, which is 1 liter equals 1000 milliliters, and I get my answer in milliliters. This is the final answer.
18:24
So, what I've done, determined how many moles of OH- how many H+, how many perchloric acid units do we need, that number of moles, and then we convert it through the molarity into volume. And that's the final answer. Typical question, make sure it's not a hang-up for you.
18:51
The redox. Redox reactors. The last topic, and clearly there's sometimes a bunch of steps you have to take, as long as you take those steps consistently, you'll be in good shape. I know people are, you know, dressing with
19:08
the homework a little bit, and stretching their heads over, and it doesn't make any sense. I don't want that out. Well, typically there's small mistakes. Either you overlooked the charge, you added one electron as opposed to two, or you forgot
19:23
that the hydrogens actually have a plus charge, and you have protons on one side, and you have a plus charge. So usually mistakes find such details. And you can get them out of the way by practicing a couple of times to make sure that you get these down.
19:44
Here's an example. Solid sulfur reacts with sodium nitrate in an acidic solution, and it forms sulfur dioxide gas, and nitrogen monoxide gas. I have to find the balanced equation. Through half the atoms. So let's do that.
20:03
So I'm going to write down just what I read. Solid sulfur, that's this guy, sodium nitrate, this is the nitrate, the sodium has left out because the sodium doesn't partake in anything. Because if you look at the products, it forms sulfur dioxide gas and nitrogen monoxide, it doesn't
20:21
even appear as a product. Of course it's there, because it doesn't do anything to the spectroscopy. I can leave the sodium ion out. Solid sulfur, nitrate, anions, product number one, sulfur dioxide, product number two, nitrogen monoxide. So this is just writing down the questions next.
20:42
Now I have to look at what is taking place. Let's look at the oxidation states of the elements. Solid sulfur oxidation state zero because that's the elemental form of sulfur it has no charge, this is zero. It has oxidation states here.
21:02
I have an oxygen here, it's bound to nitrogen. Oxygen must be minus two, okay, so that's minus two. This guy has three, that's a total of minus six, so it contributes the whole thing has to add up to minus one that means this nitrogen here must be plus five.
21:20
On the other side, this oxygen here, again, two minus two of these guys, that is four minus, that means that this guy is four plus. This oxygen again is two minus and this nitrogen there, four, must be two plus. Because the whole thing is neutral. This whole compound is neutral.
21:40
This is not a polyatomic anion so this whole thing is neutral, they must add up to zero. Okay. So what is the oxidation that's taking place? Which element is oxidized? The oxygen doesn't change. It's minus two here, minus two there, minus two there. The only thing that's changing is the sulfur and the nitrogen.
22:03
The sulfur goes from zero to plus so that one is losing electrons. That's the oxidation step. Okay? So let's write that down. The oxidation step is sulfur goes to this. Now let me try to balance this right away.
22:22
And it's having an acidic solution, so I can use H-buses and water molecules from the solution. Let's do that, and I need that, because look, there is two oxygens here, nothing here, I have to add two water molecules. On this side. In doing so, I introduce four H's, and they appear as an H-plus on the other side. Four of them. So four H-buses.
22:49
Okay, so now the elements seem to be balanced, all right, but the charges are not. They're not balanced, because there's four plus charges here, it is zero overall here, that means I have to add electrons. Four and four.
23:06
Now, there is an extra check that you can do. An extra check that you can do. Namely, this oxidation state changes from zero to plus four. It loses how many electrons? Four electrons. Ta-da! There you go.
23:26
They roll out very nicely. They roll out nicely, by the way, I just wanted to tell you, this is just, you know, they roll out nicely if the stoichiometric coefficient is one. If the stoichiometric coefficient is one and two here and there,
23:42
that means also two in front of the electrons, and it becomes eight. So, be careful with that. As long as it is one, the number of electrons you generate is going to change the oxidation state of that element. Let's look at the reduction step. The reduction step must be the nitrogen element.
24:04
So it's changing its oxidation state from plus five to plus two. Let's write it down. Nitrate on one side, on the other side, nitrogen on the other side. There we go. Let's balance this. First, the number of elements, the nitrogen is balanced. So that's good.
24:21
Oxygen is not. I need two bars on the right side. On the right-hand side. In doing so, again, I have introduced four hydrogens, they appear in the form of H+, on the reactant side in this case, and in doing so, I have balanced all the elements.
24:41
I have not solved the problem of the charges, because this zero charge overall here, 1-4+, there's a total of 3+, it needs to be zero, that means I have to add three negative charges, three electrons. Okay? Quick check, once again, three electrons, does that make sense? Yes. That makes sense. Plus five, plus two, the difference is three.
25:06
And the total of the coefficients is one here, so there should be a three appearing right there. An actual check you can apply, you don't have to, but you can train yourself to look out for those little opportunities to check on yourself.
25:21
Okay. So now I have balanced both of these guys, but now I don't have to do one another. So I have to multiply the first one by three to get 12 electrons, and the second equation by four to get 12 electrons here as well. I had four here originally, I had three here, so I multiply this whole thing by four, this equation by three, and that
25:44
allows me to add up those two equations and cross out the number of electrons. So if I do that, I get 12 electrons. So if I do that, if I add these guys up, I can cross this one out, I can cross this one out as well. So what I see is this
26:01
sulfur four times, a nitrate anion, four H+, there's three SO2s, four nitrogen monoxides, and two water molecules. Including their phases. You can quickly check yourself if everything is okay, you can count the number of elements, there's three of
26:23
three solvers here, three solvers there, there's four hydrogens, four hydrogens here, and the number of oxygens also complies. Finally, you want to check the charges. Zero charge here overall, plus four,
26:40
four minus, also zero right there. This must be right. You can check yourself, you can convince yourself that your answer is correct. So when it appears correct, from all kinds of angles, you don't have to look at it anymore, you can move on to another question.
27:04
Why don't I put the sodium ions in there? Because they are spectator ions, you can add them if you want, but they don't partake in the reaction whatsoever. If you add them in the beginning, if you add them on the reactant side, you also have to add loose sodium ions on the product side to balance it out.
27:26
In this question there are spectators. Because the question says the products are sold for the oxide and nitrogen monoxide gas, it doesn't say anything about sodium. The only thing a sodium can do is something non-specific to sit there. Yes?
27:45
Yes. You have to subtract the waters if they appear on both sides. Do they appear? There's no water on the left-hand side, so I don't have to do anything. There's also no H-plus on this side, so this is fine. okay? But yes, if water appears on this side and this side, you cross them out in such a way that the
28:05
number of water molecules that appear double is taken out. If you had four here and two there, those two go away and two are left on the other side. Now let me give you some examples that were taken from actual exams
28:22
that your predecessors wrestled with. So here we go. this is a this looks a lot like a precipitation reaction to me two solutions, a typical kind of question solution number one what is given is a volume and a molarity
28:42
and solution number two, again, a volume and a molarity you put those two things together and something is going to participate. The first part of the question is, and this is again very difficult, is you have to write down the reaction. And this is something you should be able to do. okay?
29:01
first recognize what calcium bromide is and be able to write down beryllium sulfate, and then identify the precipitate and balance the equation. So here it is. they want you to do this in the complete ionic form calcium bromide fully dissolves into ions, so it forms calcium 2+, and two bromine ions
29:26
beryllium always has charge 2+, sulfate always 2-, so there's one beryllium, there's one sulfate what is going to precipitate here? well, it is not beryllium bromide, okay, because that dissolves very well
29:42
but the sulfate and the calcium they form a non-resolving pair, and so this is the precipitate. Calcium sulfate is the precipitate, and that's why it has an acid. The beryllium and the bromine are still happy in the solution.
30:00
this is the proper way of writing it down. okay? With the phases, with the charges, ions that are ions have charges. Ions have charges. Some people tend not to write the charges on an ion. I don't know why that is. But an ion is an ion because it has a charge. If you don't write the charge, it is not an ion. It is a neutral compound. Which is something completely different. So write those charges.
30:29
Okay. Then, again, a quintessential question for you. How many grams of the precipitate in the form? I have to know how many grams of this stuff do I have. What I have to do is the terminal limiting reagent, take the
30:45
limiting reagent, mount the moles, convert it to the amount of moles of the product, and then convert the moles of the product into grams of the product. That's the step. First step, first order of business, is determining the limiting reagent. Is it calcium bromide or is it beryllium sulfate? That's the question. How to do it? Compare the actual mole ratio with the balanced mole ratio.
31:10
Okay. The balanced mole ratio is this guy, calcium bromide, over beryllium sulfate, and that is one-to-one. Okay? That's what the reaction says. The reaction says for each calcium bromide, you need one beryllium sulfate.
31:26
The fact that there's fluid in ions doesn't matter. This is still calcium bromide, this is beryllium sulfate, and they react one-to-one. So the balanced ratio is calcium bromide over beryllium sulfate equals one.
31:42
The actual ratio comes from this, right? The actual number of moles of calcium bromide over the actual number of moles of beryllium sulfate. You multiply volume by molarity, there you go, this is for calcium bromide. This thing here, the denominator in this case, is beryllium sulfate.
32:00
That ratio turns out to be .60. This is less, this is less than one, this one over here, that means that this numerator is the limiting reagent. Calcium bromide is the limiting reagent. I usually write this, LR, which is limiting reagent.
32:24
Okay. So take the limiting reagent, the amount of moles of limiting reagent, just calculate right here, that's where it is. This is the number of moles of calcium bromide convert that into the number of moles of the product. How?
32:40
Through the mole ratio. For each one mole of calcium bromide, I have one mole of calcium sulfate. So that's one-to-one. Now I know the number of moles of calcium sulfate, which is the product. All you have to do is multiply that by the molar mass of it to find the number of grams. Okay? So this is the molar mass of calcium sulfate, the product, and I find, in this particular case,
33:04
14 grams of calcium sulfate. Why is it 14? Two sig figs? Because look, it's two sig figs here. All multiplication, so I need two sig figs, the answer is, therefore, 14 grams of calcium sulfate. This is not written in scientific notation, though, right?
33:23
That's not a major, that's not a major offensive mistake. For numbers like 14 or whatever, you don't have to do that. There's a number that is really big that you are advised to write in scientific notation. Okay. Look at the way this is set up.
33:41
Bellus ratio, actual ratio, limiting reagent, mole of limiting reagent, ratio, molar mass, done. It's clear, it's easy, this helps actually get a grade of a lot, but giving you partial credit, in the case of any mistakes I'm going to give, you can still get points.
34:00
So if you can try to do it this way, again, you are enhancing your own chances of getting higher grades, because you get more partial credit. More likely to get some partial credit. Okay. Another one Right? The balanced net ionic equation, including states,
34:20
of the reaction of this. Solv magnesium hydroxide with aqueous hydrobromic acid. This is a acid-based reaction. I gave the answer straight away. This is magnesium hydroxide it is reacting with hydrobromic acid, which is HBr and what they ask is the net ionic equation.
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The Br is not doing anything, it's a spectator ion, so I can leave that one out, and just to leave the H+, I have two waters that form because I have two OH-, two OH- from magnesium hydroxide. It's two H+, forming two waters, and the rest ion is magnesium 2+. The magnesium appears in this net ionic equation because it is bound to the hydroxide as a reactant.
35:11
That's why it appears. If this will be fully dissolved, if it will be fully dissolved, if this, for instance, would not have been magnesium,
35:21
but barium, barium hydroxide, fully dissolves, then you don't have to write the barium. If this is magnesium, it binds, forms a solid, you leave the magnesium in there. Second part. How many grams of this solv magnesium hydroxide do you need to completely neutralize
35:42
30 milliliters of a certain solution of hydrobromic acid. So if this is the reaction we're talking about, I have a bucket with hydrobromic acid in it, it's an acidic solution, I have to keep adding magnesium hydroxide to it till it neutralizes. In other words, there's a couple of H- in the solution, I can calculate how many
36:02
volume, volarities, I can calculate how many H+, there are, those are the H+, that have to be compensated by adding the same amount of OH- to it. Okay? So what I want to do is I want to see how many of those H+, do I have. Volume times molarity, be careful. Middle-liter right here, so this is in terms of liters, 0.030 liters
36:25
times the molarity equals this many moles of HBr each HBr delivers one H+. So if I take that number here, I can convert that into how many moles of this stuff I need
36:41
to neutralize all the H+. This is the number of H+. So what I need is for each two moles of H, I need one mole of magnesium hydroxide. One to two. Two here, one there. See that? One here, two there. So each magnesium hydroxide is two OH-.
37:02
So with one magnesium hydroxide, I can neutralize two HBr's. That's why there's a two there. That two is that stoichiometric coefficient right there, this one is the one in front here. So I'm converting the number of HBr's into the number of magnesium hydroxides.
37:22
Now I know the number of moles of magnesium hydroxide, multiply that with the molar mass, and that's the answer. 1.36 grams of magnesium hydroxide. Therefore, 1.36 grams of magnesium hydroxide is the right answer, and the full amount of points in this case.
37:47
All right. Very good. Let me give you an example of a typical redox fraction in the exam. Here it is. Copper sulfate is reacting with solid zinc in a zinc solution that forms a
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new solution with zinc sulfate and solid copper. Okay. So now the first step I have to do is to write down the full reaction and identify the oxidation state of each element. Each atom. Here it is. Copper 2 sulfate, there it is, copper 2 sulfate, copper 2 plus, and sulfate 2 minus,
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so this is copper sulfate. Zinc, solid zinc, it forms zinc sulfate, and it forms copper. Solid copper. Zinc, solid copper. Zinc, solid copper. has oxidation state 2+. Oxidation state of zinc is almost, in all cases, 2+.
38:44
Okay, so let's assign oxidation states this copper has 2+, it says it right here, okay? this oxygen is always 2- this sulfur here, 6+, why?
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because this whole thing, sulfate, is 2- and 5- there's four oxygens Times minus two is minus eight, the whole thing has to be minus two. That means that the sulfur must be plus six. Zinc, of course, is zero. Zinc here is two plus.
39:22
And the sulfate hasn't changed. So the oxidation state is completely the same. In other words, this sulfate anion is just a spectator. It doesn't do anything. The copper has changed to oxidation state zero. That's what it says. Because it's solid copper, that means elemental, that means zero.
39:42
So this is the right way to write it down. One way, and one thing I want to point out, is that some people say, okay, there's four oxygens here, so I have to write here minus eight above the oxygen. No. We'll do that. The convention is you write the oxidation state off the element. Irrespective of how many you have.
40:04
If you're asked things in this fashion, write down the oxidation state of the atom, you have to write down the oxidation state of the element, don't multiply that with four. I know there's a four. I want you to tell me what the oxidation state of the atom is.
40:23
Okay. Find the two balanced half-reactants in this case. Well, this one is not even that complicated I recognize immediately that this sulfate anion is not really playing any role. It doesn't change its oxidation state. So I don't have to miss it necessarily.
40:41
It's fully aqueous, so I don't really need it. It's not solid. I can just go like this. Zinc is oxidized because it's losing electrons, it goes from 0 to 2+, it actually gives up to electrons, the copper takes those electrons, okay, it goes from 2+, to 0, and becomes solid copper. The sulfate has no role to play, and it can conveniently leave it out. I don't need it.
41:06
So this is the oxidation half-reaction this is the reduction step. Once you know that, you can just add these guys up, just cross out these electrons, and you find the following situation. Zinc, copper, Zinc 2+, and copper S.
41:25
So again, put those physical states always, I know that the sulfate doesn't always do it, the homework doesn't request you to put the states in it, but I'm kind of very unfortunate, so it leads to bad practice. Please make sure you add the states. Every time
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you write something, you add the states and make things multimeter, and it's easier for yourself in the long run. Okay. Actually, this equation didn't really turn out to be that terribly difficult. Any questions, folks?