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

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

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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
- Lesson Overview
- Molarity
- Molarity Cont'd
- 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
- 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
- Stoichiometry and Double-Displacement Precipitation Reactions
- Stoichiometry and Neutralization Reactions
- Stoichiometry and Neutralization Reactions
- Stoichiometry and Acid-Base Standardization
- The Experimental Setup
- Stoichiometry and Acid-Base Standardization
- Summary
- Sample Problem 1: Stoichiometry & Neutralization
- Sample Problem 2: Stoichiometry

- 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

### General Chemistry Online Course

I. Basic Concepts & Measurement of Chemistry | ||
---|---|---|

Basic Concepts of Chemistry | 16:26 | |

Tools in Quantitative Chemistry | 29:22 | |

II. Atoms, Molecules, and Ions | ||

Atoms, Molecules, and Ions | 52:18 | |

III. Chemical Reactions | ||

Chemical Reactions | 43:24 | |

Chemical Reactions II | 55:40 | |

IV. Stoichiometry | ||

Stoichiometry I | 42:10 | |

Stoichiometry II | 42:38 | |

V. Thermochemistry | ||

Energy & Chemical Reactions | 55:28 | |

VI. Quantum Theory of Atoms | ||

Structure of Atoms | 42:33 | |

VII. Electron Configurations and Periodicity | ||

Periodic Trends | 38:50 | |

VIII. Molecular Geometry & Bonding Theory | ||

Bonding & Molecular Structure | 52:39 | |

Advanced Bonding Theories | 1:11:41 | |

IX. Gases, Solids, & Liquids | ||

Gases | 35:06 | |

Intermolecular Forces & Liquids | 33:47 | |

The Chemistry of Solids | 25:13 | |

X. Solutions, Rates of Reaction, & Equilibrium | ||

Solutions & Their Behavior | 38:06 | |

Chemical Kinetics | 37:45 | |

Principles of Chemical Equilibrium | 34:09 | |

XI. Acids & Bases Chemistry | ||

Acid-Base Chemistry | 43:44 | |

Applications of Aqueous Equilibria | 55:26 | |

XII. Thermodynamics & Electrochemistry | ||

Entropy & Free Energy | 36:13 | |

Electrochemistry | 41:16 | |

XIII. Transition Elements & Coordination Compounds | ||

The Chemistry of The Transition Metals | 39:03 | |

XIV. Nuclear Chemistry | ||

Nuclear Chemistry | 16:39 |

### 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 is*0046

*going to be to double replacement precipitation reactions, acid-base neutralization, and finally*0053

*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

*Remember how we tackled dimensional analysis?--you always start with the unknown.*0359

*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

*Let's go ahead and move on; how about volume?*0468

*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

*We are not going to start with that because that is a ratio already.*0488

*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 M _{1}V_{1} equals to M_{2}V_{2}, where M_{1} is initial molarity and V_{1} is the initial volume.*0649

*M _{2} is final molarity; V_{2} 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 V _{1} which is liters, look what I get on M_{2}V_{2}.*0701

*M _{2} 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; M _{1}V_{1} equals to M_{2}V_{2}.*0740

*Let's see what we can plug in now; M _{2}V_{2} is already given to us.*0748

*M _{2} is going to be the desired concentration of 0.67 molar.*0752

*V _{2} is the desired volume; that is going to be 250 milliliters or 0.250 liters.*0758

*The M _{1} is given to us also; it is the 1.2 molar stock solution.*0769

*This V _{1} 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

*V _{1} 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 to*0819

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

*V _{final} minus V_{initial} 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

*The good news about this is it is going to be all familiar.*0891

*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 PbCl _{2} for every 1 mole of PbCl_{2}.*1041

*When all is said and done, we get 11.1 grams of PbCl _{2}.*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 of*1090

*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

*That is the nice thing about this chapter, about this lecture.*1254

*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 MgOH _{2} aqueous.*1805

*That is going to go ahead and form H _{2}O liquid and MgCl_{2} 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 MgOH _{2}.*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 MgOH _{2}.*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 including*1986

*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, Na _{2}SO_{4} 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 is*2250

*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

0 answers

Post by Evan Wang on May 18, 2017

M1V1=M2V2 looks like that physics equation but with mass and velocity. Are the two based off of the same comment?

1 answer

Last reply by: Professor Franklin Ow

Thu May 28, 2015 12:28 PM

Post by Mikayla McNulty on May 13, 2015

For question #9 shouldn't 32.10 mL be 0.0321L not 0.321L?

1 answer

Last reply by: Professor Franklin Ow

Mon Nov 3, 2014 10:59 PM

Post by Saadman Elman on October 31, 2014

Diego Robledo already addressed it before but he didn't give the whole picture. There is a mistake in example no. 9 (Stochiometry and Acid base Standardization) The mistake is you said 0.011 moles of Mg(OH)2 is produced. If you do the calculation again you will see that the answer is actually 0.107 moles. Even though it is a minute mistake at the beginning it made a huge difference in the end. Because later on we had to find the molarity which is moles divided by liter. If you divide .107 divided by .0250 you will get 4.3 Molarity and NOT 0.43 Molarity. You got .43

But overall, the lecture was awesome! I benefited from your lecture a lot. And i did great in the class. Sometimes even my professor took some notes from me when he was teaching ionization energy. He was really impressed. I can't thank you enough! You really made me change my major! I listened to all of your lectures!!

1 answer

Last reply by: Professor Franklin Ow

Mon Nov 3, 2014 10:59 PM

Post by Minky benedikt on October 23, 2014

Hi professor,

In example 5, can you please write the full equation to solve for V1?

1 answer

Last reply by: Professor Franklin Ow

Mon Oct 20, 2014 12:31 AM

Post by Joseph Patrick Balao on October 19, 2014

Hi Professor,

How do you calculate this problem

How many milliliters of 0.250 M hydrochloric acid are required to neutralize a solution containing 0.863 g barium hydroxide ?

1 answer

Last reply by: Professor Franklin Ow

Wed Jul 2, 2014 11:51 AM

Post by brandon joyner on July 1, 2014

Why is 1.2 M stock solution M1 and not M2

0 answers

Post by Saadman Elman on May 29, 2014

It's really helpful! Thanks a lot!

1 answer

Last reply by: Professor Franklin Ow

Sun Mar 16, 2014 12:26 AM

Post by Meredith Roach on March 9, 2014

On sample problem #7, how is it possible that the LR is NaCl when Pb(NO3)2 actually yields less product? To calculate Theoretical Yield, we would have to use the reactant that produces the smallest amount of product, no?

2 answers

Last reply by: Saadman Elman

Fri Oct 31, 2014 9:55 PM

Post by Diego Robledo on November 17, 2013

For Example #9, shouldn't the moles for Mg(OH)2 be 0.11 instead of 0.011? In addition, shouldn't 25.00mL Mg(OH)2, when converted to L, be 0.2500L instead of 0.02500L?

1 answer

Last reply by: Professor Franklin Ow

Thu Nov 7, 2013 5:17 PM

Post by HARRISON IGWE on September 8, 2013

calculate the volume of hydrogen produced when 6g of magnesium reacts with excess dilute hydrochloric acid at s.t.p

1 answer

Last reply by: Professor Franklin Ow

Thu Nov 7, 2013 5:19 PM

Post by HARRISON IGWE on September 8, 2013

calculate the mass of calcium oxide residue gotten when 82g of calcium carbonate is heated to a steady mass