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### Stoichiometry II

• Stoichiometry uses coefficients from a balanced chemical equation as a conversion factor to relate any (2) reactants and/or products.
• Mole to mole ratios are central to any stoichiometry problem.
• The limiting reagent or reactant dictates how much product is expected to form (known as the theoretical yield).
• Molarity expresses solution concentration, and can be used as a conversion factor.
• Acid-base titrations are used to determine (or standardize) the concentration of an unknown solution.

### Stoichiometry II

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
• Lesson Overview 0:10
• Molarity 1:14
• Solute and Solvent
• Molarity
• Molarity Cont'd 2:59
• Example 1: How Many Grams of KBr are Needed to Make 350 mL of a 0.67 M KBr Solution?
• Example 2: How Many Moles of KBr are in 350 mL of a 0.67 M KBr Solution?
• Example 3: What Volume of a 0.67 M KBr Solution Contains 250 mg of KBr?
• Dilutions 10:01
• Dilution: M₁V₂=M₁V₂
• Example 5: Explain How to Make 250 mL of a 0.67 M KBr Solution Starting From a 1.2M Stock Solution
• Stoichiometry and Double-Displacement Precipitation Reactions 14:41
• Example 6: How Many grams of PbCl₂ Can Form From 250 mL of 0.32 M NaCl?
• Stoichiometry and Double-Displacement Precipitation Reactions 18:05
• Example 7: How Many grams of PbCl₂ Can Form When 250 mL of 0.32 M NaCl and 150 mL of 0.45 Pb(NO₃)₂ Mix?
• Stoichiometry and Neutralization Reactions 21:01
• Example 8: How Many Grams of NaOh are Required to Neutralize 4.5 Grams of HCl?
• Stoichiometry and Neutralization Reactions 23:03
• Example 9: How Many mL of 0.45 M NaOH are Required to Neutralize 250 mL of 0.89 M HCl?
• Stoichiometry and Acid-Base Standardization 25:28
• Introduction to Titration & Standardization
• Acid-Base Titration
• The Analyte & Titrant
• The Experimental Setup 26:49
• The Experimental Setup
• Stoichiometry and Acid-Base Standardization 28:38
• Example 9: Determine the Concentration of the Analyte
• Summary 32:46
• Sample Problem 1: Stoichiometry & Neutralization 35:24
• Sample Problem 2: Stoichiometry 37:50

### Transcription: Stoichiometry II

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is going to be our second lecture on stoichiometry.0003

Let's go ahead and look at the lesson overview.0011

Last time we laid out a pretty good foundation for how to tackle stoichiometry problems.0014

We are just going to now build upon that foundation and apply stoichiometry to more specific cases.0022

These cases include the following; we are going first talk about solutions and their concentrations.0029

We are going to work with a lot of terminology today such as molarity and dilutions.0038

After discussing molarity and dilutions, the next specific application of stoichiometry is0046

going to be to double replacement precipitation reactions, acid-base neutralization, and finally0053

something we have not talked about yet which is called acid-base standardization via a titration.0062

After that we will go ahead and do our summary followed by our sample problems.0067

Moving on now, molarity.0077

We tend to think of a solution in chemistry as composed of two items--typically a solute and a solvent.0079

The solute is going to be present in the smaller quantity.0089

The solvent is going to be present in the major amount.0095

Of course for general chemistry lecture and for laboratory, the typical solvent of course is going to be water.0100

We always like to express a solution's concentration--basically how much solute we have relative to solvent.0111

There is a lot of different units we can use.0119

But the most common one we are going to use is called molarity.0121

The equation for molarity is defined as the following.0125

Molarity is equal to moles of solute for every liter of solution.0128

From this equation of course, we should always be very comfortable with solving for a single variable.0136

We can solve for liter of solution.0144

That is going to be basically moles of solute over the molarity.0148

Of course we can also solve for moles of solute.0157

Moles of solute is just going to be equal to molarity times liters of solution.0160

Once again you want to definitely have these equations committed to memory, especially for general chemistry.0172

Let's go ahead and jump right into a calculation.0181

How many grams of potassium bromide are needed to make 350 milliliters of a 0.67 molar KBr solution?0186

When we verbalize capital M which stands for molarity, this is verbalized as molar.0196

This is the language; this is the terminology that we use; 0.67 molar potassium bromide solution.0203

This is a first example.0210

We are just going to plug everything we know into the equation for molarity and solve for the unknown.0212

The molarity is given to us; that is 0.67 molarity.0217

That is equal to moles of solute divided by liters of solution.0223

We don't know what the moles of the solute is, the moles of KBr.0234

That is going to be what we are trying to find.0238

The liters of solution, you are actually told that in the problem.0241

It is here; it is 350 milliliters, which is 0.350 liters.0243

When we go ahead and solve for moles of potassium bromide, we get 0.2345 moles of KBr.0251

Of course you see that the question is not asking for moles, but the question is asking for grams.0262

Do you recall how to go from moles to grams?0268

What is the conversion factor called?--it is molar mass.0271

0.2345 moles of KBr times something over something.0276

That is going to give us the grams of KBr.0284

The molar mass is approximately 119 grams over 1 mole.0289

We are going to get an answer of approximately 28 grams of KBr.0294

We are really playing on the concept of the mole very heavily.0301

Once again what we learned last time in stoichiometry is going to serve as our foundation for the rest of this lecture.0306

So very important that we have those fundamentals down; let's go ahead.0312

What we just did was we used the molarity equation directly, moles over liters.0319

We just solve for the unknown.0323

But there is another way of using the molarity equation.0325

You see that the equation for molarity is basically a ratio of two units.0329

If you recall, I have said this many times before, that when you have a ratio of two units,0333

you can get a conversion factor that can be used in dimensional analysis.0338

Let's go ahead and use this approach now.0345

You see this is the exact same problem as before.0347

How many moles of KBr are in 350 milliliters of a 0.67 molar potassium bromide solution?--and how many grams is this?0350

What is unknown?--what is the only thing that is given to us that is not part of a ratio yet?0366

350 milliliters is by itself; that looks pretty good.0372

0.67 molar, remember this is a ratio of two units.0374

This is really 0.67 moles for every 1 liter.0380

Or this is going to be 1 liter over 0.67 moles.0386

Remember this is how we use molarity as a conversion factor.0392

It is going to be moles over liters or liter over mole.0395

Let's start with what is given to us that is not part of a ratio.0401

That is going to be 350 milliliters.0404

You see that in the question for molarity, the volume is actually liters.0409

What do you think we have to do to 350 milliliters?0414

We have to get it to liters.0417

That is going to be 0.350 liters times something over something.0418

Let's go ahead and use our conversion factor.0425

We want liters downstairs; we want moles upstairs.0428

That is going to be 0.67 moles for every 1 liter.0432

That is going to give us our answer of 0.2345 moles of KBr.0437

From there you can go ahead and get the 28 grams of KBr on your own.0445

Again all this is is this is the exact same problem we just did about a minute ago.0450

But all we are doing is now we are using molarity as a conversion factor instead of just using the equation itself.0455

Either way, whatever is more comfortable for you, please use that way.0463

What volume in milliliters of a 0.67 molar KBr solution contains 250 milligrams of KBr?0472

We see molarity here; we know that molarity is 0.67 moles over 1 liter.0480

The only other value is right here, 250 milligrams.0491

Remember what we have always done in stoichiometry.0496

We always wanted to get to moles; remember grams to moles.0499

That was so typical of being the first step in all of our problems, this lecture and last lecture.0506

Anytime you see mass in a stoichiometry problem, we are always going to get it to grams and then to moles.0515

0.250 grams of KBr, we are going to go ahead and get this to moles of KBr.0522

That is going to be times 1 mole divided by roughly 119 grams of KBr.0532

That gets us moles of KBr.0541

Now that we are in moles, we can use molarity as a conversion factor once again; the 0.67 moles over liters.0544

This time to get cancelled goes moles; liters goes upstairs; it is 0.67 in 1 liter.0552

That gives us our answer in liters after all is done and cancelled.0560

Now I want to get to milliliters of course; times 1 milliliter over 10-3 liters.0564

This is going to be an answer of 3.1 milliliters.0572

Once again any stoichiometry problem, always get to moles.0578

That always should be your first step.0588

You are going to hear me say that quite often.0593

Once again for stoichiometry, the first step is to get anything you can into moles.0596

The next application of stoichiometry, and this comes directly from molarity, is the practice of what we call the dilution.0604

When you order a chemical from a company, the chemical typically comes in the bulk form.0616

It is typically very concentrated; this is what we know as the stock solution.0623

When you receive the stock solution, you typically dilute it down to a desired concentration.0629

Being able to do a accurate and precise dilution is quite important in the laboratory setting.0636

To perform a dilution, the equation here is going to be used.0645

It is basically M1V1 equals to M2V2, where M1 is initial molarity and V1 is the initial volume.0649

M2 is final molarity; V2 is the final volume.0659

When you do a dilution, you typically just add water.0667

For example, sometimes parents when they give apple juice to their children, they typically add water to dilute it down.0671

Notice though that the only thing that they change is the amounts of solvent.0677

They do not change the amount of juice or sugar.0681

This amount of solute remains the same; amount of solute does not change in a dilution.0685

We can prove this mathematically.0698

Because when I take molarity which is moles over liters, and I multiply it by V1 which is liters, look what I get on M2V2.0701

M2 is moles over liters; and times liters; so moles is equal to moles.0710

Once again in a dilution, the moles of solute before equals the moles of solute after.0716

We are now going to get into a typical dilution problem.0726

Explain how to make 250 milliliters of a 0.67 molar potassium bromide solution starting from a 1.2 molar stock solution.0731

Let's go ahead and write out the equation again; M1V1 equals to M2V2.0740

Let's see what we can plug in now; M2V2 is already given to us.0748

M2 is going to be the desired concentration of 0.67 molar.0752

V2 is the desired volume; that is going to be 250 milliliters or 0.250 liters.0758

The M1 is given to us also; it is the 1.2 molar stock solution.0769

This V1 therefore is the unknown; it is pretty simple to solve for this algebraically.0776

After we are going to do this, we are going to then discuss the physical significance of this because that is another issue.0785

V1 is going to be 139.6 milliliters; what is the physical significance of this volume?0790

This is the amount of the stock solution you are actually going to take; amount of stock solution to dilute.0801

But how much water do you add then to dilute it?0815

If we want 250 milliliters as the final volume, and we are going to0819

use 139.6 milliliters of the stock, all I have to do is subtract.0827

Vfinal minus Vinitial is going to give me the volume of water to add to perform the dilution.0837

That is going to be equal to 110.4 mL of water.0850

If you were to do this physically in the laboratory, you will literally take 139.6 milliliters of stock solution.0858

To it, you would add 110.4 milliliters of water.0868

That is going to give you 250 milliliters of the 0.67 molar potassium bromide solution.0871

Once again this is how you do a standard typical dilution problem.0878

Let's move on to the next application of what we learned last time in stoichiometry.0885

This is stoichiometry and double displacement precipitation reactions.0893

Before we tackled the following questions.0900

How many grams of lead chloride can form from 50 grams of NaCl?0903

We know how to do this.0908

We went from grams of A to moles of A; then to moles of B; then on to grams of B.0911

Remember this was what we called a mass to mass conversion; mass to mass conversion.0921

Now we are going to change it up a little.0932

All we are doing is we are asking the same problem where now we are going to incorporate molarity.0934

Now how many grams of lead(II) chloride can form from 250 milliliters of 0.32 molar NaCl?0939

Remember what is always the first step in a stoichiometry problem is to get to moles; get to moles.0949

We have 250 milliliters; we have 0.23 molar; we have a volume and a molarity.0962

We are going to take 0.250 liters; we are going to multiply that by something over something.0969

That is going to give us moles of sodium chloride.0975

Liters goes downstairs; moles goes upstairs; that is going to be 0.32 over 1.0980

That is going to give us moles of sodium chloride.0988

Now what we want to do again is grams of A to moles of A, to moles of B, to grams of B.0993

We are still going to follow that flow chart.1002

This is really our moles of A.1005

Now we are going to get to moles of B.1009

We are going to take x moles of NaCl.1011

Remember how we get to the mole to mole ratio?1017

It comes from the balanced chemical equation; 2 sodium chlorides to 1 lead(II) chloride.1019

That is going to be times 1 mole of lead(II) chloride for every 2 moles of sodium chloride.1026

That is moles of B; now from moles of B, we go on to grams of B.1037

Multiply by the molar mass of lead(II) chloride which is 278.1 grams of PbCl2 for every 1 mole of PbCl2.1041

When all is said and done, we get 11.1 grams of PbCl2.1051

This is a very typical problem; don't forget what we did here was the following.1058

Basically volume times molarity gets you the moles.1064

I cannot underscore this enough of how often you are going to use this when you do a stoichiometry problem.1069

Remember volume times molarity is equal to moles.1077

It is going to be typically your first step.1080

Let's do another problem.1088

How many grams of lead(II) chloride can form when 250 milliliters of1090

0.23 molar sodium chloride and 150 milliliters of 0.45 molar lead(II) nitrate mix?1096

You notice that you are given amounts of both reactants instead of just one.1104

Remember what type of problem this is anytime you are given amounts of both reactants?1111

That is right; it is a limiting reactant problem.1116

The limiting reactant problem that we learned in the previous lecture can easily still apply to what we learn today.1119

Let's go ahead and do it.1126

Remember we are going to use what was called the smaller quantity method to solve limiting reactant problems.1128

Let's go ahead and do this.1141

Remember we are going to determine the amounts of product that can form from each of the reactant's amounts.1142

0.250 liters, this is going to be times 0.32 moles over liter of sodium chloride.1149

We want to get to lead chloride; remember it is a 1:2 ratio.1161

We just did this; that is going to give us 0.040 moles of lead(II) chloride.1164

I am going to repeat the process now for the other reactant amount.1176

0.150 liters times 0.45 moles of lead(II) nitrate over 1 liter of the lead(II) nitrate.1180

When you look back at the balanced chemical equation, the mole to mole ratio between lead(II) chloride and lead(II) nitrate is 1:1.1197

Times 1 mole of lead(II) chloride divided by 1 mole of the lead(II) nitrate.1205

When that is all calculated, you get 0.68 moles of lead(II) chloride.1214

In this specific example, the theoretical yield should be 0.040 moles of lead(II) chloride.1221

That translates to 11.1 grams of lead(II) chloride.1228

Therefore the limiting reactant in this specific example is sodium chloride.1232

Again we are using previously learned material in today's lecture; that is all.1244

It is just a slight twist on things, but you are totally capable of doing this.1249

Again it is just using old material.1252

We are really building upon what we have learned just recently.1257

The next specific application of stoichiometry is going to be to again another aqueous reaction that we have already talked.1265

That was the acid-base neutralization reaction.1274

If everybody recalls, in a neutralization reaction, it is always between an acid and a base.1277

The two products are always going to be water and a salt.1283

Again a neutralization reaction between a Bronsted-Lowry acid and a base is just another double replacement reaction.1288

This equation is balanced; so we are good to go and ready to use it.1297

How many grams of sodium hydroxide are required to neutralize 4.5 grams of hydrochloric acid?1301

Once again grams of A to moles A, goes on to moles of B, goes on to grams of B.1307

This is a typical mass to mass conversion again.1318

Don't get thrown off by the word neutralize.1321

It is just another term there that means it is going to react with each other.1324

All again this is some mass to mass conversion.1328

4.5 grams of hydrochloric acid times 1 mole divided by 36.458 grams of hydrochloric acid.1332

That is its molar mass roughly; that gives us the moles of A.1343

Moles of A to moles of B, that is going to be a 1:1 ratio.1349

That is 1 mole of sodium hydroxide over 1 mole of the HCl.1352

Finally now we are in moles of B.1359

Let's go ahead and move on to grams of B.1362

That is going to be times 39.998 grams of sodium hydroxide over 1 mole.1364

4.9 grams of sodium hydroxide are required for this reaction to proceed, are required to neutralize the indicated amount of hydrochloric acid.1372

We are going to tackle the same problem.1386

But now we are going to apply molarity to it.1388

Because this is aqueous after all; so we can use molarity anytime.1391

How many milliliters of 0.45 molar sodium hydroxide are required to neutralize so much of HCl?1395

Remember what is always the first step?1404

It is to get anything you can into moles.1406

For sodium hydroxide, all I am given is the molarity.1410

I cannot get that into moles because that says moles over liters.1412

But you see that combination again for hydrochloric acid, the volume and molarity.1416

That is always a one-two combo.1421

Anytime you see molarity and volume, multiply the two together.1424

You are going to get moles; don't forget that.1428

You are going to hear me say that again.1432

Don't forget that; this is so repetitive.1433

You can totally feel confident in doing this because it is the same tools over and over again.1436

0.250 liters of the hydrochloric acid times 0.89 moles of hydrochloric acid over 1 liter of hydrochloric acid.1443

That is moles of A; now from moles of A on to moles of B.1456

That is a 1:1 ratio from the balanced chemical equation.1460

Times 1 mole of sodium hydroxide divided by 1 mole of the hydrochloric acid.1464

That gets us... you know what?--we have to do one more step.1471

We still have to use the molarity there; my apologies.1484

Moles of sodium hydroxide goes downstairs; liters goes upstairs.1490

That is going to be 0.45; and 1 on top.1493

When all is said and done, we get 0.494 liters of sodium hydroxide required.1497

The question is asking for milliliters.1503

That is going to be 494 milliliters of sodium hydroxide required.1505

This is a nice cumulative problem because it really utilizes the concept of molarity as a conversion factor. not once but twice.1511

Again this is stoichiometry and neutralization reactions.1526

The next specific application of stoichiometry pertains also to acid-base chemistry.1531

This is what we call the concept of standardization.1538

Standardization is basically determining the concentration of an unknown.1542

To do this, we use the process known as titration.1548

Titration is a general term where a solution of known concentration is used to determine the concentration of an unknown solution.1552

In other words, we are going to standardize the unknown.1560

We are going to find out its exact concentration.1562

There are many different types of titration.1566

A very common and specific type is also acid-base titration.1568

In acid-base titration, a basic or acidic solution of known concentration is used to determine the concentration of an acidic or basic solution.1572

The analyte is what we are trying to find; this is the unknown.1585

The titrant is what we are going to be using to standardize the unknown.1594

This is the known concentration; this is the solution of known concentration.1598

Let's get into the experimental setup; a typical titration has the following.1610

A burette is basically a very precise piece of glassware used to deliver small aliquots of liquid.1615

Burettes are usually pretty precise up to 0.01 milliliters.1625

In the burette typically goes the titrant; inside the flask then goes two things.1631

It goes the analyte and also what we call an indicator.1645

What an indicator's job is is the following.1651

It is going to provide a visual sign of when the titration is complete.1653

We are going to talk about titrations, more complicated ones, down the line.1669

But for now an indicator's job again is to tell us when the titration is complete.1675

Remember this is acid-base neutralization after all.1682

This is going to tell us when the neutralization has occurred; complete neutralization has occurred.1685

The visual sign is the indicator is going to change color.1700

Again you are going to do this so much in general chemistry lab, I promise you.1709

It is one of the most traditional types of experiments we teach undergraduates.1712

Let's go ahead and jump into a typical acid-base standardization problem.1719

We have to get familiar with the terminology.1724

32.10 milliliters of 0.67 molar HCl was required to reach the equivalence point with 25 milliliters of magnesium hydroxide.1727

Determine the concentration of the analyte.1736

First let's get some terminology out of the way.1739

The equivalence point is the point in the titration where complete neutralization has occurred.1741

The point in the titration where complete neutralization has occurred.1751

This is very important that you pay attention to these key words.1764

Because that tells you pretty much how much has reacted, when the reaction has complete.1768

Step one, you always hear me... that you always want to get the moles as step one.1777

That is usually true but what is even more important before that is that we have a balanced chemical equation.1790

Anytime you are doing a stoichiometry problem, make sure you know the balanced chemical equation or get it.1797

Let's go ahead; this is HCl aqueous plus MgOH2 aqueous.1805

That is going to go ahead and form H2O liquid and MgCl2 aqueous.1815

We have to go ahead and balance it.1823

This is going to be 2 hydrochloric acids and 2 waters; why is that important?1825

Because you just know that we are going to be using a mole to mole ratio from the balanced chemical equation.1831

So it is imperative that you know the coefficients correctly.1836

Step two, get anything you can into moles; get all amounts possible into moles.1841

We have the volume of magnesium hydroxide; we can't do anything with that yet.1857

The only thing here is hydrochloric acid.1861

You see that we are given both the volume and molarity; don't forget.1865

You see that combination again; volume times molarity is going to get you moles.1869

Let's go ahead and do this; 0.3210 liters times 0.67 moles of HCl over 1 liter.1873

That gets us to moles of A.1886

Once we get in moles of A, let's go ahead and get to moles of B.1889

It is a 2:1 ratio; 1 magnesium hydroxide on top divided by 2 moles of hydrochloric acid on the bottom.1893

That is going to give us 0.011 moles of MgOH2.1905

The question is asking for the concentration of the analyte, the concentration of magnesium hydroxide.1913

The only other value in the problem we haven't used yet is the 25 milliliters.1920

Remember that we want concentration which is molarity; molarity is moles over liters.1926

We know the liters; it is given to us in the original problem.1932

There is 25 milliliters; that is going to be 0.02500 liters.1938

The moles of magnesium hydroxide we just solved for.1944

When all is said and done, we are going to get an answer of 0.43 molar MgOH2.1949

This is how you determine the concentration of unknown.1959

This is how we standardize it using the concentration of a known substance.1962

Let me go ahead and summarize today's lecture then.1969

The first take-home message is that when trying to define the concentration of a solution, the most common unit used is molarity.1972

Again molarity is moles of solute over liter of solution.1982

Molarity can be used as a conversion factor in dimensional analysis problems including1986

dilutions problems, double replacement precipitation reactions, acid-base neutralization, and acid-base standardization via titration.1991

Finally let's go over the basic steps and formats for doing general stoichiometry problems.2002

Step one is to make sure you come up with the balanced chemical equation if not provided for you already; balanced chemical equation.2009

Step two, step two is to get all amounts you can into moles; get all amounts possible into moles.2025

Remember there are some common patterns that we have seen already.2038

Usually we can go from grams to moles.2042

That is going to be via molar mass.2046

But another thing we have seen is using molarity.2050

That was volume times molarity is going to get you moles.2052

Again that is using molarity as a conversion factor.2057

The first one here we use molar mass as a conversion factor.2068

That is going to get us into moles of A.2077

Then once you are in moles of A, you can go on to moles of B.2084

From moles of B, you can go on then to solve the problem.2090

Sometimes the question asks us for grams of B.2094

Sometimes the question asks us for volume of B.2098

Again if we are going to do grams of B, once again that is going to be molar mass as a conversion factor.2104

But if we are going to go to volume of B, we are going to be using simply molarity again as a conversion factor.2111

I hope that is a nice little summary of how to tackle general stoichiometry problems.2119

Let's now work on some more sample problems.2124

Sample problem number one, how many milliliters of 0.45 molar sodium hydroxide are required to neutralize 250 milliliters of 0.89 molar sulfuric acid?2129

Step one, we don't have any balanced chemical equations.2141

Let's go ahead and come up with it on our own; sulfuric acid reacting with sodium hydroxide.2145

That is going to go ahead and give us water and our salt, Na2SO4 aqueous.2156

Let's go ahead and balance it.2165

We are going to be needing two of these and of course two of those.2168

Step one is done.2174

Step two is to get any quantities you can into moles; get into moles.2176

Remember what are the possible combinations?2185

Usually we go from grams to moles via molar mass.2189

But you don't see any grams in this problem.2192

The other alternative is to do volume times molarity getting you into moles.2194

Here we have the volume and molarity for sulfuric acid.2198

That is going to be 0.250 liters times 0.89 moles of sulfuric acid over 1 liter.2202

That gets us into moles of A; now we want to go to moles of B.2213

The mole to mole ratio is 1 sulfuric acid to 2 sodium hydroxides.2225

Times 2 moles of sodium hydroxide for every 1 mole of sulfuric acid.2229

Finally the question is asking for volume; we want to get now into milliliters.2237

To do that, we use molarity; times 1 liter over 0.45 moles.2242

That is going to give us a answer of 0.99 liters of the sodium hydroxide which is2250

going to be 990 milliliters required for neutralization to occur of the specified amount of sulfuric acid.2258

That is sample problem number one; let's now move on to sample problem number two.2269

How many grams... right away I see grams; that is mass.2273

So I know for sure that I am going to be using probably molar mass as a conversion factor.2279

Mass via molar mass as conversion factor.2285

How many grams of barium sulfate can form when so much sodium sulfate reacts with so much barium nitrate?2291

Let's go ahead; we see that we have both amounts of two reactants.2299

So you know that this is a limiting reactant problem.2306

Let's go ahead and go through it.2316

We are going to be using the smaller quantity method.2317

Step one, get the balanced chemical equation if it hasn't been provided to you already.2327

Here we have to come up with it; balance chemical equation.2333

It is going to be sodium sulfate aqueous reacting with barium nitrate aqueous.2342

That is going to go on and form barium sulfate solid and sodium nitrate aqueous.2353

Let's go ahead and balance this guy.2363

I am going to be needing two of those for our balanced chemical equation.2365

Alright, step one is complete.2370

Step two... you guys know the drill.2373

Step two is to get all quantities you can into moles; get into moles.2376

Look here; for both of the reactants, for sodium sulfate we have both the volume and the molarity.2382

For barium nitrate, we have the volume and the molarity.2390

So volume times molarity gets us into moles.2393

We are going to do this for each reactant.2398

We are going to see which one gives us the smaller theoretical yield for the product.2400

Let's work with sodium sulfate first.2407

0.250 liters times 0.32 moles of sodium sulfate over 1 liter.2409

That gets us into moles of A; now we go on to moles of B.2419

The mole to mole ratio between sodium sulfate and barium sulfate is 1:1.2423

1 mole of barium sulfate over 1 mole of sodium sulfate.2429

That is going to give us 0.080 moles of barium sulfate.2437

We are going to continue on with the smaller amount method for solving the limiting reactant problem.2444

We are going to repeat the process for the other reactant amount.2449

0.125 liters times 0.87 moles of barium nitrate divided by 1 liter.2452

The mole to mole ratio between barium nitrate and barium sulfate is also 1:1.2463

That is 1 mole, barium sulfate over 1 mole of barium nitrate.2469

That is going to give us 0.11 moles of barium sulfate.2478

Using our smaller amount method, the theoretical yield is going to be 0.080 moles of product.2488

That is going to be 18.7 grams of barium sulfate formed, which means sodium sulfate is the limiting.2496

Again this is a limiting reactant problem using molarity as a conversion factor.2511

I hope you leave today's lesson feeling pretty confident and comfortable doing a stoichiometry problem.2515

It seems like a lot of calculations.2523

But if you follow the basic steps, you will get it every time.2524

Remember step one is balance chemical equation if not given to you already.2528

Step two is going to be basically getting all the quantities you can into moles.2533

If you need to go from grams to moles, it is always molar mass.2538

If you need to go from volume to moles, it is always going to be molarity for these two lectures.2543

Thank you for using Educator.com again; I will see you again next time.2554