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Unformatted text preview: 78 Chapter 8 STOICHIOMETRY, Part 2
©2004, 2008, 2011 Mark E. Nob/e We continue with more stoichiometry topics. Our fundamentals are in place. The four basic steps
which we covered in the last Chapter remain our primary method for attacking problems of this type, at
least within the context of dimensional analysis. As food for thought, we open this Chapter with a
question many people often ask in their daily lives. 8.1 What do we do with the leftovers??? What leftovers? Well, we didn't have any yet, but you can end up with some, depending on how you
actually run a reaction. I will begin this part by referring to our H2/C|2 reaction from Chapter 7 and the
examples which we worked at that time. Recall the balanced equation. H2 + CI2 —> 2 HCI
Let me summarize those results. ./ The problem started with 24.77 9 H2 as given information. ./ We first calculated that this 24.77 9 H2 would react evenly (no more, no less) with 871.1 g Clz. ./ Secondly, we calculated that 895.9 g HCI could be made from the 24.77 g H2. ./ For a third example, we used the 871.1 g Cl2 to calculate the amount of product, which gave us
the same answer of 895.9 g HCI. If we actually do the reaction beginning with 24.77 9 H2 (given info) and 871.1 g Cl2 (calculated by
stoichiometry), then we say we are running the reaction at "stoichiometric amounts" or "stoichiometric
conditions". The two phrases are equal and you can use either one. The terms mean that all reactants
are present in amounts as calculated by stoichiometry for each other. When our reaction of 24.77 9 H2
and 871.1 g Cl2 is done, you should have 895.9 g HCI, as calculated. There would only be HCI molecules
afterwards; all H2 molecules and all Cl2 molecules reacted. There would be no more molecules of H2 left
and there would be no more molecules of Cl2 left. Frequently we don't do reactions at stoichiometric conditions. Let's say we load the pot with 24.77 9
H2 and 914.3 g Clz. Notice that this is more Cl2 than we can use. When the reaction finishes, you would
still have 895.9 g HCI, but you also have leftover chlorine. The 24.77 g of hydrogen molecules still
reacted with just 871.1 g of chlorine molecules, as required by the stoichiometry and the balanced
equation. That part doesn't change. But you put more than enough chlorine molecules into the pot to
start. How much more? You started with: 914.3 g of chlorine molecules
and the reaction used: 871.1 g of chlorine molecules
which means there are: 43.2 g of chlorine molecules leftover What happens to the leftovers? Nothing. They're still there at the end. When the reaction is over, you
have your product of 895.9 g of HCI molecules and you have 43.2 g of chlorine molecules left over. There
are no H2 molecules since all of these reacted. Why would you want to put too much of one reactant into the pot? Isn't that wasteful? Well, maybe,
but not necessarily. There are many reasons for doing this and it's a very common practice in the
laboratory and in industry. Having extras of one reactant often helps to make the whole process go better
and faster. How? Well, I can't go into that right now because we haven't covered enough other turf.
Much of this ties into "equilibrium" which will get a bit of mention in Chapter 12 and much more extensive
coverage in much later chapters. Some of the reasons tie into the speed of reactions which will be
discussed beginning in Chapter 48. Some of this ties into still other aspects. We're not really in a spot
where I can go into details. That means you'll have to take my word for it. There are some things,
however, which we can take care of right now, and that is why I even brought this up. We will see how
this is handled by stoichiometry calculations. In this present example starting with 914.3 g chlorine molecules, we knew there was going to be
leftover chlorine because we had done all the calculations in Chapter 7 for stoichiometric conditions. We
knew that 24.77 9 H2 and 871.1 g Cl2 were the stoichiometric amounts for each other because we
calculated one from the other. However, a typical problem of this type does not start there. I'll show you
how these typical problems start, but first we need some terminology. Chapter 8: Stoichiometry, Part 2 79 8.2 Excess and limiting We don't call leftovers "leftovers". We call them excesses. The leftover amount is called the "excess
amount". In the illustration above starting with 24.77 9 H2 and 914.3 g Clz, we had 43.2 g leftover Clz.
We say 43.2 g Cl2 is the excess amount. We also say Cl2 is the "excess reagent" for the reaction. The
excess reagent is the one whose initial amount is greater than needed based on the amounts of other
reactants. On the other hand, H2 is the reactant which runs out completely. When all 24.77 g of
hydrogen molecules reacted, that was the end of it. No more H2 molecules, no more reaction. We call
H2 the "limiting reagent". The limiting reagent is the one in shortest stoichiometric supply at the start of
the reaction. When it's gone, it's gone. The reaction is done. Since the limiting reagent runs out and
ends the reaction, the limiting reagent determines how much product can be made. It doesn't matter if
other reactants are still floating around ready for action. ONCE ANY REACTANT IS USED UP, YOU'RE
FINISHED. Done. Completed. Over. That's all the product you can get. The limiting reagent limits the amount of product possible. CAUTION!!! It is important for you to be aware of the difference between this scenario and a reaction
conducted under stoichiometric conditions. Under stoichiometric conditions, gﬂ reactants are put into the
pot in stoichiometric amounts. They aﬂ run out and are used up at the same time. Nobody is in excess.
Since they all run out evenly and at the same time, the amount of product can be calculated from a_ny of
the reactants. We actually did this in the last Chapter with the HZ/Cl2 reaction. For the HCI product, we
calculated that 895.9 g HCI could be made from the 24.77 9 H2 and we also calculated that the same
895.9 g HCI could be made from the 871.1 g Clz. Let me repeat: under stoichiometric conditions the amount of product can be calculated from any of the reactants. On the contrary, if any reactant is
limiting, then the amount of product must be calculated from that amount of that reactant. OK, let's turn now to a typical problem with limiting reagent. By the way, this kind of problem is
actually called a "limiting reagent problem". I'm getting tired of H2 and CIZ. Let's do something different.
We'll do a reaction to make silver bromide. We'll start with silver nitrate and sodium bromide as the
reactants. Sodium nitrate will also be produced by this process. AgNO3 + NaBr —> AgBr + NaNO3 Ever use silver nitrate? Well, maybe I should ask have you ever had it rubbed on your body? Aqueous
solutions of silver nitrate are used as a disinfecting agent in medical procedures such as minor operations,
oral surgery, etc. If you had it rubbed on your skin, you would know. Within hours, your skin would turn
much darker and it would stay that way for days until it wears off. Speaking of medical things, I can also
mention sodium bromide. Sodium bromide and potassium bromide had been used as sedatives for many
years through much of the early 1900's. (It was even in the movies, as a spellbound Alfred could tell
you.) This use has been discontinued due to side effects, but these compounds are still used in veterinary
applications. What about the product, silver bromide? Ever use it? Have you ever taken a photograph
with film (not digital)? That's rare nowadays but it was the norm not too long ago. Silver bromide was
the leading light—sensitive compound used in photographic films. It's the stuff that actually reacts with
light. Here's the setup for the problem. An aqueous solution of 14.01 g silver nitrate and an aqueous
solution of 10.91 g sodium bromide are prepared in separate beakers. Both compounds are ionic
compounds. They are colorless by themselves and when they dissolve in water their solutions are also
colorless. When the two solutions are mixed, a reaction occurs which causes a cloudy, yellowish-white
appearance. The cloudiness is due to tiny particles of solid silver bromide which eventually settle to the
bottom of the container. The sodium nitrate product is colorless and stays dissolved in the water. By the
way, I mentioned the use of phases in chemical equations in Chapter 6. This would be a nice illustration
to show you how "aqueous" phase is used. AgNO3(aq) + NaBr(aq) —’ AgBr(s) + NaNO3(aq)
Don't forget: aqueous means dissolvedin water. Since the AgBr isn't dissolved, it can't be called aqueous. OK, that's the setup. Now the question: for this process with the given amounts of reactants, how
many grams of silver bromide can be made? First, a bit of background. Whenever the amounts of two or more reactants are given to you in a
problem, then the very first thing you need to know is whether these are stoichiometric conditions or not.
Sometimes the problem will say so. Sometimes, like right now, it won't. If it doesn't say so, you should
assume it is NOT stoichiometric conditions. Why is this necessary? If the reactant amounts are
stoichiometric amounts, then you can calculate the amount of product from any one of the reactant 80 Chapter 8: Sto/chiometry, Part 2 amounts. On the other hand, if the reactant amounts are not stoichiometric amounts, then you must
calculate the amount of product based on the limiting reagent. We started with 14.01 9 silver nitrate and 10.91 g sodium bromide. We assume these are not
stoichiometric amounts. We assume that one reactant is the limiting reagent and the other reactant is
the excess reagent. Which is limiting and which is excess? That's not obvious. You can't just look at the
number of grams because the answer also involves moles. There are several ways to do these kinds of
problems and I will illustrate one which is fairly common. This method applies to those problems which
ask for the amount of a product as part of the problem. There are other ways which your instructor may
cover and prefer. Again, these are multiple methods. Use the one that works best for you. The present approach is based on a sentence from upstairs: the limiting reagent limits the amount
of product possible. We can calculate the amount of product which would be possible from the given
amount of each reactant. That means we do a separate stoichiometry calculation starting from each
reactant amount. Then we compare those results. The reactant which will make the least amount of
product is the limiting reagent and that determines the amount of product. Let's get started. The problem specifically asks forthe amount of silver bromide which can be prepared; sodium nitrate
will also be made, but the problem does not request that. We ignore sodium nitrate and we focus only
on the silver bromide product. We will do two calculations: we will calculate how much silver bromide can
be made from 14.01 9 silver nitrate and we will calculate how much silver bromide can be made from
10.91 g sodium bromide. AgNO3 + NaBr -+ AgBr + NaNO3
14.01 g 10.91 g ??? 9 Our problem begins with these two basic stoichiometry calculations. Step 1, as always, needs a balanced
equation and this has been provided. You can check this yourself. (Don't always assume that equations
are balanced when given to you. Sometimes instructors want you to balance the equation first.) First calculation: how many 9 AgBr can be made using 14.01 g AgNO3?
This is standard stoichiometry. Plot your path: start with the amount of AgNO3 provided.
0 Step 2. Remember the molar mass connection to get to moles. The molar mass for AgNO3 is 169.9 g.
g AgNO3 -' mol AgNO3 mol AgNO3
169.9 g AgNO3 - Step 3. Rxn ratio. The coefficients in the balanced equation for AgNO3 and AgBr are 1:1.
9 AgNO3 -' mol AgNO3 -’ mol AgBr 1 mol AgBr
1 mol AgNO3 - Step 4. Now molar mass for AgBr. It's 187.8 g.
g AgNO3 -’ mol AgNO3 -* mol AgBr -' g AgBr Of the two possible molar mass conversion factors, you need: Of the two possible rxn ratios, you need: Of the two possible molar mass conversion factors, you need: mol AgBr
- Now string it all together.
path: 9 AgNO3 -’ mol AgNO3 -' mol AgBr -’ g AgBr
14.01 g AgNO3 X mol AgNO3 X 1 mol AgBr X 187.8 g AgBr = 15.49 g AgBr
169.9 g AgNO3 1 mol AgNO3 mol AgBr This ends our first calculation. We have determined that 15.49 g AgBr product can be made from the
14.01 g AgNO3 which were provided. Chapter 8: Stoichiometry, Part 2 81 Now we need to do the second calculation, basing it on the given amount of NaBr. Plot your path.
It's basically the same. You'll need the molar mass of sodium bromide (102.89 g), along with other
numbers from above. You have to do more of the details here, since I'm not giving all of them for every
problem. I'll give you the full path. You fill in what's needed. path: 9 NaBr -* mol NaBr —’ mol AgBr ** g AgBr 10.91 g NaBr x x x ——-————————- = 19.91 g AgBr I gave you the answer so you can be sure you put all the conversion factors in correctly. You need to fill
this in. Plug in your numbers, punch them out, round them off, and be sure you get 19.91. If not, find
and fix your mistake. Alright let's summarize what we have from the calculations and we can make our conclusions. I First calculation: We have enough AgNO3 to make 15.49 g of formula units of AgBr.
./ Second calculation: We have enough NaBr to make 19.91 g of formula units of AgBr. Remember. The limiting reagent limits the amount of product. The calculations show that a smaller
amount of product is obtained from the AgNO3. We see that we can make 15.49 g AgBr, after which all
AgNO3 has reacted. At that point, there are no more formula units of AgNO3 available in order to
continue. The reaction is over. Done. Although the second calculation says we have enough NaBr to
make 19.91 g AgBr, the reaction still stops at 15.49 g AgBr clue to the limit imposed by the amount of
AgNO3. Thus, we have too much NaBr and there will be leftovers of NaBr. We conclude AgNO3 is the
limiting reagent. NaBr is the excess reagent. The answer to the original question is that 15.49 g AgBr
can be made under these conditions. You're done with the problem. Now let's change the problem and carry it out a bit further to illustrate several aspects. We've just
concluded that you end up with 15.49 g AgBr in the pot. You also end up with some NaNO3 as the other
product. How much NaNO3 do you get? The original problem did not ask for this, but you can calculate
the amount of this product, too. Fortunately, you don't have to do two calculations again. That's because
the limiting reagent applies to the WHOLE reaction. As soon as the limiting reagent is gone, everything
stops. Once the limiting reagent is identified, all stoichiometric outcomes for all other reagents are
determined from that. Since we already know that AgNO3 is the limiting reagent, we can calculate the
amount of NaNO3 just from 14.01 g AgNO3. As above, I'll give the full path and I'll give you the answer.
You can go back in and insert the conversion factors and make sure you're doing it right. You'll need the
molar mass for sodium nitrate (85.00 9), along with some earlier numbers. path: gAgNO3 -' molAgNO3 -’ molNaNO3 -’ gNaNO3 14.01 g AgNO3 x x x = 7.009 g NaNO3 Be sure you get this answer. Otherwise, find and fix the error. Here‘s another aspect. Sodium bromide is the excess reagent, which means that some of it will be
left over. How much NaBr is actually left over? In order to answer that, we must first calculate how much
of the NaBr actually reacted. Since the limiting reagent determines the stoichiometric outcome for
everybody else, that's what you start with: your first job is to calculate the amount of NaBr which reacted
with the 14.01 g AgNO3. I'll outline it and you can fill it in. path: gAgNO3 -’ molAgNO3 -’ molNaBr -’ gNaBr 14.01 g AgNO3 x ———-——~ x x = 8.484 g NaBr This result says that, out of the 10.91 g sodium bromide which were provided initially, 8.484 g actually
reacted with the 14.01 9 silver nitrate. To find the amount of NaBr left over, we just do a subtraction the
same way as we did for the Hz/Cl2 example earlier in this Chapter. 82 Chapter 8: Stoichiometry, Part 2 You started with: 10.91 g of formula units of NaBr
and the reaction used: 8.484 g of formula units of NaBr
which means there are: 2.43 g of formula units of NaBr leftover There you have it: 2.43 g NaBr is the excess amount.
Let me summarize everything we did. > We started with: 14.01 g AgNO3 10.91 g NaBr
> We calculated the amount of
AgBr product possible from each: 15.49 g AgBr 19.91 g AgBr > The reaction is limited to the lesser amount of AgBr; here, the lesser amount derives from the AgNO3,
so AgNO3 is the limiting reagent. 15.49 g of AgBr product were possible, and this ended the problem
as originally stated. > We took the problem further and we calculated that 7.009 g NaNO3 would also be made, based on
AgNO3 as limiting reagent. > Also based on AgNO3, we calculated that 8.484 g of the original NaBr would actually react in the
process. > Finally, we showed that 2.43 g NaBr would be left over as the excess amount.
The grand tally in the pot after the reaction is done would be: 15.49 g of formula units of silver bromide product
7.009 g of formula units of sodium nitrate product
2.43 g of formula units of sodium bromide leftover
no formula units of silver nitrate at all That's everything from the reaction. These types of problems can get very long and drawn out. Think how long the problem would take
if there were three or more reactants! Yes, this can happen. These can be a lot of work, but keep in mind
that most of the grunt work is the basic stoichiometry string. As always, be able to think through the given information and figure out where you have to go. The
bottom line comes down to the usual thing: practice, practice, practice. 8.3 The real world versus the ideal world I want to go into something different at this point. You could say that it's a different aspect of
stoichiometry but it's really a dose of reality. I mentioned reality checks as Catch #3 in Chapter 1. I said
we would run into them. This is our first, good reality check. As described in Chapter 1, reality checks occur when things don't go by plan. In the real world of
doing chemical reactions and making compounds, there are numerous problems which can arise along
the way. Some reactions require heating and sometimes you might not heat it hot enough or you might
not heat it long enough. Some reactions require cooling and then similar considerations can apply as to
whether you got it cold enough or cooled it long enough. Many things can also happen even after the
reaction is done. The primary goal after many reactions are done is to get the product you want
separated from everything else. I can illustrate this with the AgNO3/NaBr reaction which we just finished.
I said we wanted to make silver bromide. Our starting point was a solution of silver nitrate and a solution
of excess sodium bromide. We mix the two solutions and we make a slurry. There's the solid AgBr which
you want, there's dissolved NaNO3 which was the other product, there's the excess amount of NaBr
reactant (still dissolved), and there's the water in which you did the reaction. OK, now you made the
silver bromide and now you have to get it out of the pot. Since the AgBr formed as a solid and everything
else stays dissolved in solution, then we can filter the mixture to collect the AgBr. Then we rinse it with
clean water and then we dry it out. These steps are easy. Things can be much worse. What if two solids
are formed and they're mixed together? What if the product you want is a liquid? What if the product
you want is a gas? For some reactions there could be 5 - 10 or more steps involved. By the way, all of
these steps after a reaction is done are called the "work-up". Spills and other losses can occur anywhere along the road. If you've ever done lab experiments you
may know about some of this. One thing students dread about lab class is winding up with Mr./Ms. Klutzo
as their lab partner. They drop things, spill things, make a mess everywhere, etc. Each step in the Chapter 8: Stoichiometry, Part 2 83 process is an invitation for something to go wrong. Actually, anybody can have an accidental drop or spill.
Some people are more likely than others. And anybody can have just a plain old, bad day. This stuff
happens. I've been there. Notice the parallels to cooking. Baking needs to be done at the right temperature for the right
amount of time or the cake is flat. Setting a gelatin mold needs chilling for a certain amount of time or
you wind up with slop. Spills can happen. Maybe you just cranked that old mixer up to warp speed and
half of your cake batter is on the wall trying to blend in with the wallpaper pattern. How about cookies?
Probably one of the best series of kitchen chemistry reactions, in my opinion, is based on the Original
Nestlé Toll House Chocolate Chip Cookies recipe. It says you should get 5 dozen. I've done this recipe
on several occasions (years ago). I never got 60 cookies. Of course, there are special rules which apply
for cookies, since cookies require multiple trays in and out of the oven in order to do the whole batch.
There‘s a separate Law of Nature that applies to multi—tray baking operations; that Law of Nature states
that, in any multi-tray baking operation, one tray must burn. You can't get around this. It doesn't matter
if you preheat the oven for thirty minutes or thirty days. One tray (or more) will burn. Some of the
outcome will not be the product it's supposed to be. It won't be edible. At least not by humans. All of these things illustrate a fundamental outcome in cooking and in chemistry in general. There's
the theoretical amount of product which you should get and then there's the actual amount of product
which you really end up with after all reaction and work-up steps are done. This is a simple case of ideal
versus real. This is the reality check. There are several new terms to introduce. The theoretical or ideal amount of the product which you
should get is called the "theoretical yield". In baking, the recipe tells you what it should be. In chemistry,
the theoretical yield is the amount of product as calculated by stoichiometry. In the chocolate chip cookie
recipe, the theoretical yield is 60 cookies based on using the recipe amounts. In the reaction to make
silver bromide, the theoretical yield is 15.49 g AgBr based on the limiting reagent of 14.01 g AgNO3. The actual or real amount of the product which you finally end up with is called the "actual yield".
This is something that happens; this is not something you can predict. For example, if you happen to
wind up with 45 cookies, then that is your actual yield. If you happen to wind up with 13 g AgBr, then
that is your actual yield. A third term is "percent yield". This is just the ratio of real versus ideal written as a percentage. percent yie|d = ___a._c_tua|—ylelg___ x 1000/D theoretical yield
For an actual yield of 45 cookies, your percent yield is calculated as follows. percent yield = W x 100% = 75%
60 cookies For an actual yield of 13 g AgBr, your percent yield is calculated as follows. percent yield = w x 100% = 84%
15.49 g AgBr Percent yield is a very useful measure of the overall success of the entire operation as a whole. That
includes the reaction itself and all the work—up steps associated with getting the actual product out. We
want a percent yield as close as possible to 100%, but sometimes you're stuck with lower and sometimes
you're stuck with much lower; it depends on how hard the reaction and the work-up are to do. In
research, each time we publish a new compound we have to report how we made it, which includes the
reaction and the work-up steps. In my own research group over the years, we've reported percent yields
for different procedures from ~10% up to ~90%. Why the range? It just depends. And there's no sure
way of predicting it in advance. So keep these terms in mind. Theoretical yield, actual yield and percent yield. Remember that
theoretical yield is ALWAYS the amount calculated by stoichiometry. And that means based on limiting
reagent, too. Whenever a stoichiometry problem asks you to calculate the amount of product possible
from some amount of such-and-such, then it's asking for the theoretical yield. 84 Chapter 8: Stoichiometry, Part 2 Problems 1. True or false.
a. A limiting reagent is used up completely during a reaction.
b. Some amount of an excess reagent will be present at the end of a reaction.
c. If a reaction is done at stoichiometric amounts of all reactants, then no reactant is in excess. 2. The following equation is balanced.
S4N4 + 4AgO —. 4AgS + 4N0 The reaction is conducted beginning with 26.37 g S4N4 and 69.23 9 A90. How many grams of NO
can be made? 3. The following equation is balanced.
PCI3 + 3 HF —. PF3 + 3 HCI
The reaction starts with 29.6 g PCI3 and 23.4 g HF.
a. How many grams of PF3 can be made?
b. How many grams of HCI can be made? c. How many grams of the excess reagent are left over at the end? 4. The following equation is balanced.
2 KOH + H2504 -+ KZSO4 + 2 H20
The reaction is conducted using 6.21 g KOH. The actual yield of KZSO4 is 9.22 g. What is the percent
5. The following equation is balanced.
C6H6 + 9 CI2 -» CGCI12 + 6 HCI The reaction is conducted beginning with 4.608 g C6H6 and 40.25 g Clz. The actual yield of HCI is
11.4 9. What is the percent yield? ...
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This note was uploaded on 01/12/2012 for the course CHEM 201 taught by Professor Hoyt during the Fall '07 term at University of Louisville.
- Fall '07