For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

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### Stoichiometry Examples

- The Mol is the central unit in Chemistry, and every Stoichiometry problem involving a reaction will involve a Mol Ratio between species.
- You may work from the Molecular, Total Ionic or Net Ionic equations, but Net Ionic is often the best.
- Before any Stoichiometry problem BALANCE your equation. The coefficients give you the all-important Mol Ratios.

### Stoichiometry Examples

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Stoichiometry Example 1 0:36
- Example 1: Question and Answer
- Stoichiometry Example 2 6:57
- Example 2: Questions
- Example 2: Part A Solution
- Example 2: Part B Solution
- Example 2: Part C Solution
- Example 2: Part D Solution
- Stoichiometry Example 3 17:56
- Example 3: Questions
- Example 3: Part A Solution
- Example 3: Part B Solution
- Example 3: Part C Solution

### AP Chemistry Online Prep Course

### Transcription: Stoichiometry Examples

*Hello, and welcome back to Educator.com.*0000

*We are going to continue our discussion of AP Chemistry.*0003

*We just finished talking about precipitation reactions, acid-base reactions, and oxidation-reduction reactions.*0006

*Today's lesson is going to consist of just some general stoichiometry problems, with respect to these three classes of reactions, just to get us really, really, really comfortable with the idea of dealing with the chemistry first, and then doing the mathematics.*0012

*This is the heart and soul of chemistry.*0030

*Let's just jump in and see what we can do.*0033

*OK, our first example is going to be Example 1: How many grams of calcium hydroxide are required to neutralize 35 milliliters of a 0.15 Molar nitric acid solution?*0037

*Nitric acid is HNO _{3}.*0079

*So, we have this calcium hydroxide, and we have this 35 milliliters of a .15 Molar acid solution, and we're going to drop in some calcium hydroxide.*0081

*We want to know how many grams we actually need to drop in there to completely neutralize.*0092

*When they say "neutralize," they are talking about the neutralization reaction.*0097

*You notice, you have an acid, and you notice, you have a hydroxide; so we're talking about H ^{+} + OH^{-} forming water.*0101

*Again, it's 1:1; 1H + 1OH forms 1 water.*0108

*Well, let's just write the equation and work from there.*0114

*Again, the idea is: don't just jump into the math--this is chemistry.*0116

*The idea is: write an equation--that is the first thing you want to do; write an equation, balance it, do anything else you need to do to the equation (maybe there are two or three different equations--maybe you want to break it up into a net ionic--something like that); whatever works for you, but make sure you understand the chemistry first.*0121

*The math, again, is incidental; it's something that you will need, but if you don't understand the chemistry, you will never be able to do the math, because you won't know where to go in the problem.*0138

*So, let's do the equation: we have calcium hydroxide (Ca(OH) _{2}), and we're going to add that to an HNO_{3}.*0147

*That is going to form...so there is going to be--this is an acid-base neutralization, so outside, you're going to end up forming water (which I write as HOH, and I just do that because, when I'm balancing, I like H on one side, OH on the other; it's just easier for me; plus, it reminds me that H and OH are actually separate--that the two H's are not...) plus calcium nitrate (CaNO _{3}, and I believe this is a 2).*0159

*Let's balance this off: let's put a 2 here; let's put a 2 here; and there you go: 2 H's, 2 H's; 2 nitrates, 2 nitrates; 1 calcium, 1 calcium; 2 hydroxides, 2 hydroxides.*0192

*Good; so, this says that, for each mole of calcium hydroxide, I'm going to neutralize 2 moles of nitric acid.*0205

*Let's just jump right on in!*0217

*Let me see: they say "How many grams of calcium"...so here, we're going to need to go from mol of 35 milliliters of a .5 molar HNO _{3}, so we're going to go from moles of HNO_{3}...*0221

*Using a mole ratio, we're going to find moles of Ca(OH) _{2}, and then grams of the Ca(OH)_{2}.*0236

*So, this is just a simple stoichiometry problem--two conversions; great.*0247

*Let's start by finding the number of moles of HNO _{3}.*0254

*We have 0.035 liters, because it's 35 milliliters, times 0.15 mol per liter.*0257

*I personally prefer to work one thing at a time; you are welcome to do multiple conversions right on one line; I prefer to do it one at a time, again, simply because I like to see the chemistry at each step.*0274

*It's a personal choice; I'll let you handle it as you please.*0287

*When I do this one, it tells me that I get 0.00525 moles of HNO _{3}.*0291

*Well, 0.00525 mol of HNO _{3}; the mole ratio is 2 moles of HNO_{3} per 1 mole of calcium hydroxide.*0303

*1 mole of calcium hydroxide...therefore, when we cancel that, we end up with (actually, in this case, let's just go ahead and do the straight conversion so you can see it, and then one mole of calcium hydroxide...)--the molar mass is 74.08 grams, and when we do that times that divided by this, we should end up with 0.194 grams of calcium hydroxide.*0329

*That's it; so, if I have 35 milliliters of a .15 Molar nitric acid solution, I need to drop in .194 grams of calcium hydroxide to completely neutralize the acid.*0368

*In other words, what I'm adding here is a base--hydroxide--to H ^{+}, the hydrogen ion, which is acid, to neutralize it to form water under this reaction.*0383

*This is a standard neutralization reaction: acid + base goes to water, and the salt is calcium nitrate.*0399

*Acid + base goes to salt and water--always, always, always.*0407

*Let's go to Example #2.*0412

*Example #2 is the kind of an example that is a little bit intimidating to some students, and the reason it's intimidating is not because the problem is hard; it is because the problem can be kind of long.*0423

*When kids see a bunch of words--the longer the paragraph, or the longer the word problem, they tend to get intimidated by the length.*0436

*So, psychologically, it tends to sort of interfere; just read the problem!*0445

*You will discover that the problem, more often than not, is actually quite simple.*0449

*A lot of times, you will have very, very short word problems that are quite complicated.*0453

*So, don't let the length of the problem actually scare you.*0457

*All right, so a sample of iron ore is dissolved in acid to convert the iron metal to iron 2+ ion.*0461

*The reason it does that is because acid has a higher reduction potential than iron.*0491

*So, when you drop iron in acid, hydrogen gas will bubble off.*0495

*Iron will oxidize to iron ion; hydrogen ion will reduce to hydrogen gas.*0500

*OK, now the sample is then titrated with 57.50 milliliters of a 0.0344 Molar permanganate ion solution.*0507

*The redox equation for this (in other words, the oxidation-reduction--redox is just short for oxidation-reduction), for the titration, is: 1 mole of MnO _{4}, plus 5 iron 2+ ions, plus 8 hydrogen ions, goes to manganese 2+ ion plus 5 moles of iron 3+ (notice, iron went from 2+ to 3+), plus 4 waters.*0544

*This is an oxidation-reduction reaction; don't let it scare you.*0586

*All that it is saying is that, once I have taken the ore and dropped it in some acid, the acid has converted the iron to iron 2+; and now, I'm going to take that iron 2+, and in order to find out how much is actually in there, I'm going to be converting it to...oxidizing it with permanganate...I'm going to be converting it to iron 3+.*0589

*In the process, I'm going to produce manganese ion and some water, and it's going to take place in an acidic medium.*0611

*That is all that is going on here; there is nothing strange about it.*0616

*So, here are some of the questions that we can ask about this--really easy questions, actually.*0622

*How many moles of permanganate were added to the solution?*0631

*That is the first question.*0645

*B (the second question--B): How many moles of Fe ^{2+} were in the sample that was titrated?*0647

*C: How many grams of iron metal were in the ore?*0666

*And, last but not least: If the ore weighed 0.785 grams, then what was the percent by mass (or the mass percent) of iron?*0679

*It's a long problem--actually very, very straightforward: you just read it right off.*0707

*You are given the equation right there; next it's just basic stoichiometry.*0711

*Let's start: so, how many moles of permanganate were added to the solution to convert all of the ion?*0720

*Well, all right, let's move forward here.*0726

*Let's see: we have 0.5750 liters (because that is 57.50 milliliters) times 0.0344 moles per liter.*0737

*That is what they are saying; they want to know how much permanganate was added--well, you take the number of liters that you add times the molarity, and that will give you the number of moles that was added.*0758

*You end up with 0.001955 mol of MnO _{4}^{-}; that's it; that's A--nice and simple.*0768

*B: they want to know how many moles of iron 2+ were in the sample that you titrated.*0784

*Well, you had the equation; the equation is: 1 mole of permanganate reacts with 5 moles of iron ion.*0794

*Therefore, we just do a simple mole ratio: 0.001955 mol of MnO _{4}^{-} times 5 mol of Fe^{2+} per every 1 mole of MnO_{4}^{-}; units of cancellation; and you end up with 0.009775 mol of Fe^{2+}--very, very nice, see?--nothing particularly complicated about it.*0803

*Well, now they want to know what the mass is; now that you have the moles of the ion, all of the ion came from the metal, so 0.009775 mol of Fe ^{2+}, and the molar mass of iron is 55.85 (I think--yes) grams per mole.*0840

*So, we end up with 0.546 grams of iron in the sample of the ore.*0868

*Hopefully, this is pretty clear.*0879

*D: they want a mass percent; so we're going to take the mass of iron, divided by the mass of the ore, and then multiply by 100 for a percentage.*0880

*So, mass percent equals 0.546 grams divided by 0.785 grams, times 100, and you end up with 69.5%.*0890

*That means, of the ore, 69.5% of it was iron.*0909

*That is it; the stoichiometry was actually pretty straightforward.*0914

*I do want to talk a little bit--just mention again what it is that actually happened.*0918

*We took iron metal, and we reacted it with acid (let me actually write it--instead of on top of the arrow, let me write it separately); we reacted it with acid; we dropped it into acid to convert it into iron 2+ ion.*0921

*So again, hydrogen ion has a lower reduction potential than iron does; this is an oxidation-reduction reaction.*0939

*Lower reduction potential, therefore--in other words, iron is actually lower than hydrogen on the activity series, so it's going to oxidize this to iron ion.*0944

*Now, this iron ion that was oxidized--we wanted to oxidize it some more; we wanted to convert it to iron 3+, so we added some permanganate solution.*0958

*When we add the permanganate solution, we end up converting it to iron 3+, and in the process, we end up converting the manganese (well, notice: if this is going to be oxidized, something has to be reduced, so...)--look at what is happening here.*0971

*Let's go through some oxidation states.*0987

*Oxygen is 2-, always; there are 4 oxygens; therefore, there is a total of 8- charge on the oxygen.*0989

*Well, 8-...the total charge on the permanganate is -1; therefore, manganese itself has to have an oxidation state of +7.*0997

*Over here, it has an oxidation state of +2; it has been reduced.*1006

*Manganese took 5 electrons from iron; from 5 of these, it took one electron apiece in order to turn into manganese 2+, and it converted them into 5 atoms of that.*1011

*This had to take place in acidic solution, which is what that 8 H is, and in the process, it also formed 4 waters.*1026

*In other words, the oxygen had already taken electrons from manganese; manganese came into contact with iron and stole the electrons from iron.*1036

*Oxygen went over to the H's and formed water; that is what this is saying; that is all that is going on here.*1044

*Don't let all of this, everything else, just confuse you.*1050

*As long as you understand the chemistry, you can follow the chain of logic.*1054

*Everything is completely intuitive--nothing is happening here that you don't know would happen, anyway, just from your experience.*1058

*This isn't bizarre quantum mechanics kind of stuff; this is just straight things running into each other and taking things from each other.*1065

*OK, let's do another example here.*1075

*Example 3: I have: 150 milliliters of 0.25 Molar potassium hydroxide solution is mixed with 210 milliliters of a 0.17 Molar nickel sulfate.*1084

*Oops, actually, I don't need...nickel is +2; SO _{4} is...I don't really need the parentheses, although if you put the parentheses, it's not a problem; redundancy is never an issue.*1116

*It's when you leave off information--that is when the problems start.*1125

*A: We want you to write an equation for the reaction that takes place.*1129

*So now, we're not telling you what is happening; we want you to decide what is happening--if anything happens at all...what takes place?*1138

*OK, and B: How many grams of precipitate form?*1149

*This is going to be a limiting reactant problem.*1164

*C: What is the concentration of each ion left in the solution?*1167

*"Left in the solution"--at the end, once we have mixed the two solutions.*1185

*OK, well, let's start with A; so we want to write an equation for the reaction that takes place here.*1192

*We have potassium hydroxide plus nickel sulfate; let's just throw it out there: potassium hydroxide, nickel sulfate...*1198

*OK, so what is going to happen here?*1209

*It looks like you're going to end up with some kind of...potassium hydroxide is soluble, and nickel sulfate, if I'm not mistaken, is soluble; so this is a double replacement reaction--just sort of switch partners inside.*1213

*So, we'll do potassium with sulfate, and hydroxide with nickel: NiOH...this time, it's nickel hydroxide...NiOH _{2} + K_{2}SO_{4}.*1230

*Let's balance it; so we need 2 potassiums to balance the potassium; 2 hydroxide, 2 hydroxide; 1 sulfate, 1 sulfate; 1 nickel, 1 nickel.*1243

*All we needed was a 2 in front of the potassium hydroxide, and that takes care of it.*1252

*Now, nickel hydroxide is not soluble, so I'm going to draw a little down arrow, showing that it is a precipitate.*1258

*You can also put an S for solid--not a problem.*1263

*Potassium sulfate is soluble; nickel sulfate is soluble; that is soluble; so, just for the sake of practicing our net ionic--what you're going to end up with is (once you do your total ionic--I'll skip the total ionic--you can do it yourself, hopefully): nickel ion is going to react with 2 moles of hydroxide ion to produce one mole of nickel hydroxide as a solid precipitate, which we are going to find the mass of.*1267

*OK, so let's go ahead and see what we can do.*1297

*How many grams of precipitate form?--that is what we want to do next, so B.*1300

*We need to find out how many moles of hydroxide and how many moles of nickel sulfate--in particular, how many moles of nickel and how many moles of hydroxide, to see which one is the limiting reactant.*1305

*Let's go ahead and deal with this equation, since this is the chemistry.*1318

*Again, net ionic: that is your chemistry.*1327

*So, we want to find out how much hydroxide: well, it says we have 0.150 liters of a 0.25 mole per liter solution; it gives us 0.0375 moles of potassium hydroxide.*1331

*1 mole of potassium hydroxide releases 1 mole of hydroxide, so that is the same.*1354

*So, we have (oops...again, with these stray lines showing up)...this implies that we have 0.0375 mol of hydroxide ion.*1362

*I hope that makes sense: one mole of these releases one mole of hydroxide ion, because there is only one hydroxide in the compound.*1378

*OK, how about the nickel 2+?*1385

*Well, we have 210 milliliters of a .17 Molar, so 0.210 liters times 0.17 mol per liter; that gives me 0.0357 mol of nickel sulfate.*1390

*And again, 1:1...one mole of nickel sulfate releases one mole of nickel, so that implies that I have 0.0357 mol of nickel ion.*1415

*OK, and be very careful; it just so happens that these numbers were kind of crazy: you end up with .0375, .0357; be very careful which one you choose--don't confuse them.*1430

*I actually did confuse them earlier, when I was working on this.*1440

*I need to find out what the limiting reactant is, so I'm going to just pick one of them; I'm going to pick the hydroxide, 0.0375 moles of hydroxide ion.*1447

*I'm going to multiply it by the mole ratio: the hydroxide ion is...I need 2 moles of hydroxide for 1 mole of nickel reacted.*1460

*That is going to end up giving me 0.01875 mol of nickel 2+ required.*1475

*That is the calculation that I just did.*1488

*I found the number of moles of the two things involved in the reaction, and now, I just picked one of them to find out how many of the other I required.*1490

*Well, do I have .01875 moles of nickel?--yes, I do.*1498

*Because I have enough nickel, that means that the hydroxide is what is limiting.*1507

*Let me just write that down: limiting; I always like to do that, and circle it, just so I don't forget.*1511

*Because that is limiting, that is the number that I take to finish off my calculations.*1520

*0.0375 moles of hydroxide times...1 mole of nickel hydroxide for every 2 moles of hydroxide ion used; therefore, I end up producing 0.01875 moles of nickel hydroxide.*1526

*Now, I convert this to grams: 0.01875 moles of nickel hydroxide, times its molar mass, grams per mole--it ends up being 92.69, and we get 1.74 grams of nickel hydroxide precipitated.*1569

*So, we found the mass; OK, now our final is ions left over.*1603

*What is the concentration of the ions left over?*1611

*Well, let's list the ions that are left over.*1614

*We had potassium; that was one of the spectator ions--that is from the potassium hydroxide.*1620

*Hydroxide was a limiting reactant, which means it ran out completely, so there is no more hydroxide left.*1626

*However, nickel did not run out, so we have some nickel left.*1631

*And, we have sulfate.*1636

*OK, now concentration means moles per liter: so concentration--it means moles per liter.*1639

*But now, notice: we mixed two solutions; our final volume is not the same as the individual; so, we have to make sure to specify our final volume.*1651

*Volume final is equal to 0.150 liters, plus 0.210 liters, which equals 0.360 liters: now these ions are floating around in more volume, so they are more dilute.*1661

*So, let's do our potassium concentration; and concentration is listed with these little square brackets.*1679

*It is equal to...well, we said that 1 potassium hydroxide releases 1 mole of hydroxide; it releases 1 mole of potassium.*1687

*We produced .0375 moles, so the potassium number of moles is the same as the number of moles of hydroxide.*1698

*So, we have 0.0375 moles of K ^{+} over the total volume, which is 0.360 liters, and that will give us a concentration of 0.104 Molar--moles per liter; that is the potassium.*1707

*OK, now let's do the nickel.*1727

*We started off with 0.0357 moles, and we used up 0.01875 moles; so what we are left with is 0.01695 moles of nickel ion.*1731

*I hope that makes sense.*1754

*We started with a certain amount of nickel; what we used up was what was reacted with the hydroxide; so, the final amount of nickel in there is only 0.01695 moles.*1756

*We take that number, and we divide by the total volume that it is floating around in, which is 0.360 liters, and we end up with a concentration of 0.047 Molar.*1767

*I'll just write this little thing again--nickel concentration, moles per liter.*1783

*Our final one: SO _{4}^{2-}.*1790

*SO _{4}^{2-}: OK, so one molecule of (we'll do it down here) nickel sulfate releases 1 atom of nickel ion, plus one atom of sulfate.*1794

*So, the concentration of sulfate ion (it didn't participate in anything) was the original, which is this number: .0357 moles of sulfate, right?*1818

*.0357 moles of sulfate ion, divided now by the total volume, .360 liters, gives us a molarity of 0.099 Molar.*1831

*There you have it: you have a precipitation reaction; you have...any of these can happen, all in one--you can have precipitation with acid-base; you can have just straight neutralization, oxidation-reduction...*1849

*The stoichiometry is essentially just the same; what you want to do is recognize the chemistry, be able to write an equation for it, and follow the chemistry.*1864

*If you can follow what each atom is doing with what, the stoichiometry should start to make sense, eventually.*1873

*If it doesn't at this point, again: don't worry about it, because we're going to do a ton of these.*1880

*We're just going to keep doing them over and over and over again, until they are completely second nature.*1885

*As we are doing the problems, we will also be discussing theory simultaneously.*1889

*That is the best way to do it: as you do a problem, run through the theory, make sure everything makes sense, and eventually, it will all start to come together.*1893

*OK, thank you for joining us for stoichiometry problems here at Educator.com.*1902

*We will see you next time; take good care; goodbye.*1907

1 answer

Last reply by: Professor Hovasapian

Mon Jun 27, 2016 6:58 PM

Post by Jeffrey McNeary on June 26, 2016

at 20:30, how did you know to write Ni(OH)2? In other words, how did you know that a Ni ion has a 2+ charge?

0 answers

Post by Micheal Bingham on April 17, 2015

@ about the 5-minute mark, to keep the rules of significant digits should 0.15*0.035 = .0053 since we should only keep 2 significant figures? What would we use in the lab? .0053 or .00525? Thanks !

1 answer

Last reply by: Professor Hovasapian

Wed Dec 24, 2014 12:01 AM

Post by sadia sarwar on December 23, 2014

hello sir

your teaching is really helpful. thank you so much:))

sir i am stuck on one question can you please help on how to do it?

A 0.500 g sample of sodium sulfate (Na2SO4) and a 0.500 g sample of aluminium

sulfate (Al2(SO4)3) were dissolved in a volume of water, and excess barium chloride

was added to precipitate barium sulfate.

What was the total mass of barium sulfate produced?

1 answer

Last reply by: Professor Hovasapian

Tue Jun 10, 2014 8:52 PM

Post by Alice Rochette on June 9, 2014

For example 3, part a how did you know that Ni would react with (OH)2 and K2 would react with SO4?

3 answers

Last reply by: Professor Hovasapian

Sat Apr 20, 2013 6:32 PM

Post by Antie Chen on April 18, 2013

What's difference between reduction potential and standard reaction potential (E0)? Are they opposite? Higher reduction potential means lower standard reaction potential ?

1 answer

Last reply by: Professor Hovasapian

Wed Apr 10, 2013 1:50 AM

Post by Kendrick Miyano on April 9, 2013

For example 3.a, the problem is to write an equation for the reaction. For these types of problems, how do you accurately predict the outcome? I understand that there needs to be a double displacement in this case, but what I do not understand is how to come up with the compounds such as K2S04 or Ni(OH)2. Do you consider the charge of S04 and OH and then just figure out that you need K2 and Ni to balance the charges? Am I supposed to be able to answer these problems without looking at a chart that lists the charges?

3 answers

Last reply by: Professor Hovasapian

Wed May 1, 2013 4:41 AM

Post by Kendrick Miyano on April 9, 2013

I believe that for example 2.a the answer should be in fact 0.001978.

0 answers

Post by Jeffrey Herrschaft on May 20, 2012

Thanks for explaining everything very clearly. Your videos helped me pass all my Chem test and now I'm using your videos for the final as well. Your hair is awesome...kick ass chemistry!