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Lecture Comments (18)

1 answer

Last reply by: Professor Hovasapian
Mon Jun 27, 2016 6:58 PM

Post by Jeffrey McNeary on June 26 at 06:19:18 PM

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!

Related Articles:

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

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 HNO3.0079

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 HNO3.0147

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 (CaNO3, 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 HNO3, so we're going to go from moles of HNO3...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 HNO3.0254

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 HNO3.0291

Well, 0.00525 mol of HNO3; the mole ratio is 2 moles of HNO3 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 MnO4, 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 Fe2+ 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 MnO4-; 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 MnO4- times 5 mol of Fe2+ per every 1 mole of MnO4-; units of cancellation; and you end up with 0.009775 mol of Fe2+--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 Fe2+, 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; SO4 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...NiOH2 + K2SO4.1230

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: SO42-.1790

SO42-: 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