Download Stoichiometry_files/Zumdahl-Limiting Reactants

9
For Mg(OH)2:
Step 2 Using the molar mass of Mg(OH)2, which is 24.31 g ! 2(16.00 g) !
2(1.008 g) # 58.33 g, we determine the moles of Mg(OH)2 in
1.00 g of Mg(OH)2.
mol Mg(OH)2
1.00 g Mg(OH)2 " 1 ________________
# 0.0171 mol Mg(OH)2
58.33 g Mg(OH)2
# 1.71 " 10!2 mol Mg(OH)2
Step 3 To determine the moles of HCl that react with this amount of
2 mol HCl
Mg(OH)2, we use the mole ratio _______________
.
1 mol Mg(OH)2
2 mol HCl
1.71 " 10!2 mol Mg(OH)2 " _______________
1 mol Mg(OH)2
# 3.42 " 10!2 mol HCl
Step 4 1.00 g of Mg(OH)2 neutralizes 3.42 " 10!2 mol of HCl. We
have already calculated that 1.00 g of NaHCO3 neutralizes only
1.19 " 10!2 mol of HCl. Therefore, Mg(OH)2 is a more effective
antacid than NaHCO3 on a mass basis.
Active Reading Question
In the problem above, we added equal masses (1.0 g) of each antacid.
Why would one antacid consume more stomach acid than the other?
SECTION 9.2
REVIEW QUESTIONS
1 Why do we need to convert mass to moles in
stoichiometry problems?
2 Solutions of sodium hydroxide cannot be kept
for very long because they absorb carbon
dioxide from the air, forming sodium
carbonate. The unbalanced equation is
NaOH(aq) ! CO2(g) n Na2CO3(aq) ! H2O(l)
Calculate the number of grams of carbon
dioxide that can be absorbed by complete
reaction with a solution that contains 5.00 g
of sodium hydroxide.
3 Show how the steps in solving a stoichiometry problem are similar to a method that
could be used in a supermarket to determine
which of two products would be the best buy.
4 You react 10.0 g of nitrogen gas with hydrogen
gas according to the following reaction.
N2(g) ! 3H2(g) n 2NH3(g)
a. What mass of hydrogen is required to
completely react with 10.0 g sample of
nitrogen gas?
b. What mass of ammonia is produced from
10.0 g of nitrogen gas and sufficient
hydrogen gas?
5 You react hydrogen gas with chlorine gas
according to the following reaction.
H2(g) ! Cl2(g) n 2HCl(g)
What mass of HCl(g) can be produced from
2.5 " 103 g of hydrogen gas with an excess
of chlorine gas?
6 Determine what mass of carbon monoxide
and what mass of hydrogen are required to
form 6.0 kg of methanol by the reaction
CO(g) ! 2H2(g) n CH3OH(l).
RESEARCH LINKS
9.2 • Using Chemical Equations to Calculate Mass • 295
SECTION 9.3
Limiting Reactants and Percent Yield
Key Terms
• Limiting reactant
(Limiting reagent)
• Theoretical yield
• Percent yield
Objectives
• To understand the concept of limiting reactants
• To learn to recognize the limiting reactant in a reaction
• To learn to use the limiting reactant to do stoichiometric calculations
• To learn to calculate percent yield
A. The Concept of Limiting Reactants
Earlier in this chapter, we discussed making sandwiches. Recall that the
sandwich-making process could be described as follows:
2 pieces bread ! 3 slices meat ! 1 slice cheese n sandwich
In our earlier discussion, we always purchased the ingredients in the correct
ratios so that we used all of the components, with nothing left over.
Now assume that you came to work one day and found the following
quantities of ingredients:
20 slices of bread
24 slices of meat
12 slices of cheese
How many sandwiches can you make? What will be left over?
To solve this problem, let’s see how many sandwiches we can make
with each component.
Bread:
1 sandwich
20 slices bread " ________________
# 10 sandwiches
2 slices of bread
Meat:
1 sandwich
24 slices meat " _______________
# 8 sandwiches
3 slices of meat
Cheese:
1 sandwich
12 slices cheese " ________________
# 12 sandwiches
1 slice of cheese
How many sandwiches can you make? The answer is 8. When you run
out of meat, you must stop making sandwiches. The meat is the limiting
ingredient.
What do you have left over? Making 8 sandwiches requires 16 pieces
of bread. You started with 20 pieces, so you have 4 pieces of bread left.
You also used 8 pieces of cheese for the 8 sandwiches, so you have
12 $ 8 # 4 pieces of cheese left.
Active Reading Question
There are more slices of meat than of bread or cheese. So why does the
meat limit the number of sandwiches you can make?
In this example, the ingredient present in the largest number (the meat)
was actually the component that limited the number of sandwiches you
could make. This situation arose because each sandwich required 3 slices
of meat—more than the quantity required of any other ingredient.
296 • Chapter 9 • Chemical Quantities
9
A Closer Look
When molecules react with each other to form products, considerations
very similar to those involved in making sandwiches arise. We can illustrate
these ideas with the reaction of N2(g) and H2(g) to form NH3(g):
N2(g) ! 3H2(g) n 2NH3(g)
Consider the following container of N2(g) and H2(g):
= H2
= N2
What will this container look like if the reaction between N2 and H2 proceeds to completion? To answer this question, you need to remember that
each N2 requires 3 H2 molecules to form 2 NH3. To make things clear, we
will circle groups of reactants:
= H2
= N2
= NH3
Before the reaction
After the reaction
In this case, the mixture of N2 and H2 contained just the number of
molecules needed to form NH3 with nothing left over. That is, the ratio of
the number of H2 molecules to N2 molecules was
15
H2
3H
______
" _____2
5 N2
1 N2
This ratio exactly matches the numbers in the balanced equation
3H2(g) ! N2(g) n 2NH3(g).
This type of mixture is called a stoichiometric mixture—one that contains
the relative amounts of reactants that matches the numbers in the balanced
equation. In this case all reactants will be consumed to form products.
9.3 • Limiting Reactants and Percent Yield • 297
Now consider another container of N2(g) and H2(g):
= H2
= N2
What will the container look like if the reaction between N2(g) and H2(g)
proceeds to completion? Remember that each N2 requires 3 H2. Circling
groups of reactants we have
= H2
= N2
= NH3
Before the reaction
After the reaction
In this case, the hydrogen (H2) is limiting. That is, the H2 molecules are
used up before all of the N2 molecules are consumed. In this situation the
amount of hydrogen limits the amount of product (ammonia) that can
form—hydrogen is the limiting reactant. Some N2 molecules are left over in
this case because the reaction runs out of H2 molecules first.
To determine how much product can be formed from a given mixture
of reactants, we have to look for the reactant that is limiting—the one
that runs out first and thus limits the amount of product that can form.
In some cases, the mixture of reactants might be stoichiometric—that is, all
reactants run out at the same time. In general, however, you cannot assume
that a given mixture of reactants is a stoichiometric mixture, so you must
determine whether one of the reactants is limiting.
Limiting reactant
(limiting reagent)
The reactant that is
completely used up when
a reaction is run to
completion
The reactant that runs out first and thus limits the amounts of products
that can form is called the limiting reactant (limiting reagent).
To this point, we have considered examples where the numbers of reactant molecules could be counted. In “real life” you can’t count the molecules
directly—you can’t see them and, even if you could, there would be far too
many to count. Instead, you must count by weighing. We must therefore
explore how to find the limiting reactant, given the masses of the reactants.
298 • Chapter 9 • Chemical Quantities
9
Active Reading Question
What is meant by the term “limiting reactant”?
B. Calculations Involving a Limiting Reactant
Manufacturers of cars, bicycles, and appliances order parts in the
same proportion as they are used in their products. For example,
auto manufacturers order four times as many wheels as engines and
bicycle manufacturers order twice as many pedals as seats. Likewise, when
chemicals are mixed together so that they can undergo a reaction, they
are often mixed in stoichiometric quantities—that is, in exactly the correct
amounts so that all reactants “run out” (are used up) at the same time.
Let’s consider the production of hydrogen for use in the manufacture
of ammonia. Ammonia, a very important fertilizer itself and a starting
material for other fertilizers, is made by combining nitrogen from the
air with hydrogen. The hydrogen for this process is produced by the
reaction of methane with water according to the balanced equation
CH4(g) ! H2O(g) n 3H2(g) ! CO(g)
Let’s consider the question, What mass of water is required to react exactly
with 249 g of methane?
Where do we want to go?
How much water will just use up all of the 249 g of methane, leaving no
water or methane remaining?
What do we know?
• CH4(g) ! H2O(g) n 3 H2(g) ! CO(g)
•
Ammonia being dissolved in
irrigation water to provide
fertilizer for a field of corn
249 g CH4
How do we get there?
Drawing a map of the problem is helpful.
CH4(g)
!
H2O(g)
249 g
CH4
Grams of
H 2O
Use molar mass
of CH4
Use molar mass
of H2O
Moles of
CH4
Use mole ratio
from balanced
equation
n
3H2(g) ! CO(g)
Moles of
H 2O
9.3 • Limiting Reactants and Percent Yield • 299
Step 1 The equation is already balanced.
Step 2 We first convert the mass of CH4 to moles, using the molar
mass of CH4 (16.04 g/mol).
1 mol CH4
249 g CH4 ! ____________
" 15.5 mol CH4
16.04 g CH4
Step 3 Because in the balanced equation 1 mol of CH4 reacts with
1 mol of H2O, we have
1 mol H2O
15.5 mol CH4 ! __________
" 15.5 mol H2O
1 mol CH4
Step 4 Therefore, 15.5 mol of H2O will react exactly with the given
mass of CH4. Converting 15.5 mol of H2O to grams of H2O
(molar mass " 18.02 g/mol) gives
18.02 g H2O
15.5 mol H2O ! ____________
" 279 g H2O
1 mol H2O
i
nformation
The reactant that is
consumed first limits the
amounts of products that
can form.
This result means that if 249 g of methane is mixed with 279 g of water,
both reactants will “run out” at the same time. The reactants have been
mixed in stoichiometric quantities.
If, on the other hand, 249 g of methane is mixed with 300 g of water,
the methane will be consumed before the water runs out. The water will be
in excess. In this case, the quantity of products formed will be determined
by the quantity of methane present. After the methane is consumed, no
more products can be formed, even though some water still remains. In this
situation, the amount of methane limits the amount of products that can be
formed, and it is the limiting reactant.
In any stoichiometry problem, where reactants are not mixed in stoichiometric quantities, it is essential to determine which reactant is limiting
to calculate correctly the amounts of products that will be formed. This
concept is illustrated in Figure 9.1. Note from this figure that because there
are fewer water molecules than CH4 molecules, the water is consumed first.
After the water molecules are gone, no more products can form. So in this
case water is the limiting reactant.
Figure 9.1
A mixture of 5CH4 and 3H2O
molecules undergoes the reaction
CH4(g) # H2O(g) n
3H2(g) # CO(g).
Note that the H2O molecules
are used up first, leaving two
CH4 molecules unreacted.
300 • Chapter 9 • Chemical Quantities
9
E XA MP L E 9 . 7
Stoichiometric Calculations: Identifying the Limiting Reactant
Suppose that 25.0 kg (2.50 ! 104 g) of nitrogen gas and
5.00 kg (5.00 ! 103 g) of hydrogen gas are mixed and reacted
to form ammonia. Calculate the mass of ammonia produced when
this reaction is run to completion.
Solution
Where do we want to go?
How much ammonia will be produced when the reaction is complete?
What do we know?
• N2(g) " H2(g) n NH3(g)
•
•
2.5 ! 104 g N2
5.00 ! 103 g H2
How do we get there?
Step 1 The balanced equation for this reaction is
N2(g) " 3H2(g) n 2NH3(g)
This problem is different from the others we have done so far
in that we are mixing specified amounts of two reactants. To know
how much product forms, we must determine which reactant is
consumed first. That is, we must determine which is the limiting
reactant in this experiment. To do so, we must add a step to our
normal procedure. We can map this process as follows:
N2(g)
"
3H2(g)
2.50 ! 104 g
N2
5.00 ! 103 g
H2
Use molar
mass of N2
Use molar
mass of H2
Moles of
N2
Use mole ratios
to determine
limiting
reactant
n
2NH3(g)
Moles of
H2
Moles of
limiting reactant
We will use the moles of the limiting reactant to calculate the
moles and then the grams of the product.
9.3 • Limiting Reactants and Percent Yield • 301
Step 2 We first calculate the moles of the two reactants present:
1 mol N2
2.50 ! 104 g N2 ! __________
" 8.92 ! 102 mol N2
28.02 g N2
1 mol H2
5.00 ! 103 g H2 ! __________
" 2.48 ! 103 mol H2
2.016 g H2
Step 3 Now we must determine which reactant is limiting (will be consumed
first). We have 8.92 ! 102 mol of N2. Let’s determine how many moles
of H2 are required to react with this much N2. Because 1 mol of N2
reacts with 3 mol of H2, the number of moles of H2 we need to react
completely with 8.92 ! 102 mol of N2 is determined as follows:
8.92 ! 102 mol N2
3 mol H2
1 mol N2
Moles of
H2 required
3 mol H
8.92 ! 102 mol N2 ! _________2 " 2.68 ! 103 mol H2
1 mol N2
Is N2 or H2 the limiting reactant? The answer comes from the
comparison
Moles of H2
available
2.48 ! 103
less
than
Moles of H2
required
2.68 ! 103
We see that 8.92 ! 102 mol of N2 requires 2.68 ! 103 mol of H2 to
react completely. However, only 2.48 ! 103 mol of H2 is present. This
means that the hydrogen will be consumed before the nitrogen runs
out, so hydrogen is the limiting reactant in this particular situation.
Note that in our effort to determine the limiting reactant, we could
have started instead with the given amount of hydrogen and calculated the moles of nitrogen required.
2.48 ! 103 mol H2
1 mol N2
3 mol H2
Moles of N2 required
1 mol N
2.48 ! 103 mol H2 ! _________2 " 8.27 ! 102 mol N2
3 mol H2
Thus, 2.48 ! 103 mol of H2 requires 8.27 ! 102 mol of N2. Because
8.92 ! 102 mol of N2 is actually present, the nitrogen is in excess.
Moles of N2
available
8.92 ! 102
302 • Chapter 9 • Chemical Quantities
greater
than
Moles of N2
required
8.27 ! 102
If nitrogen is in excess, hydrogen will “run out” first; again we
find that hydrogen limits the amount of ammonia formed.
H2 is limiting reactant.
i
N2 is in excess.
nformation
Always check to see
which, if any, reactant is
limiting when you are
given the amounts of two
or more reactants.
Step 4 Because the moles of H2 present are limiting, we must use this
quantity to determine the moles of NH3 that can form.
2 mol NH
2.48 ! 103 mol H2 ! ___________3 " 1.65 ! 103 mol NH3
3 mol H2
Step 5 Next, we convert moles of NH3 to mass of NH3.
17.03 g NH
1.65 ! 103 mol NH3 ! ____________3 " 2.81 ! 104 g NH3
1 mol NH3
" 28.1 kg NH3
Therefore, 25.0 kg of N2 and 5.00 kg of H2 can form 28.1 kg of NH3.
The strategy used in Example 9.7 is summarized in Figure 9.2.
Grams
of H2
Molar
mass
of H2
Moles
of H2
Figure 9.2
A map of the procedure used in Example 9.7
H2
limiting
Grams
of N2
Molar
mass
of N2
Moles
of H2
2 mol NH3
3 mol H2
Moles
of NH3
Molar
mass
of NH3
Grams
of NH3
Moles
of N2
L e t ’s R e v i e w
Steps for Solving Stoichiometry Problems Involving
Limiting Reactants
Step 1 Write and balance the equation for the reaction.
Step 2 Convert known masses of reactants to moles.
Step 3 Using the numbers of moles of reactants and the appropriate
mole ratios, determine which reactant is limiting.
Step 4 Using the amount of the limiting reactant and the appropriate mole
ratios, compute the number of moles of the desired product.
Step 5 Convert from moles of product to grams of product, using the
molar mass (if this is required by the problem).
9.3 • Limiting Reactants and Percent Yield • 303
EXA MP LE 9 .8
Stoichiometric Calculations: Reactions Involving the Masses of Two Reactants
Nitrogen gas can be prepared by passing gaseous ammonia over solid
copper(II) oxide at high temperatures. The other products of the reaction
are solid copper and water vapor. How many grams of N2 are formed when
18.1 g of NH3 is reacted with 90.4 g of CuO?
Solution
Where do we want to go?
How much nitrogen will be produced when the reaction is complete?
What do we know?
Copper(II) oxide reacting with
ammonia in a heated tube
•
NH3(g) ! CuO(s) n N2(g) ! Cu(s) ! H2O(g)
•
18.1 g NH3
•
90.4 g CuO
How do we get there?
Step 1 Balance the equation:
2NH3(g) ! 3CuO(s) n N2(g) ! 3Cu(s) ! 3H2O(g)
Step 2 From the masses of reactants available, we must compute the moles
of NH3 (molar mass " 17.03 g) and of CuO (molar mass " 79.55 g).
1 mol NH3
18.1 g NH3 # ____________
17.03 g NH3
" 1.06 mol NH3
1 mol CuO " 1.14 mol CuO
90.4 g CuO # ____________
79.55 g CuO
Step 3 To determine which reactant is limiting, we use the mole ratio
between CuO and NH3 to determine how CuO is required.
mol CuO " 1.59 mol CuO
1.06 mol NH3 # 3___________
2 mol NH3
Then we compare how much CuO we have with how much of it
we need.
Moles of
CuO
available
1.14
less
than
Moles of CuO
needed to
react with
all the NH3
1.59
Therefore, 1.59 mol of CuO is required to react with 1.06 mol of
NH3, but only 1.14 mol of CuO is actually present. So the amount
of CuO is limiting; CuO will run out before NH3 does.
Note that CuO is limiting even though the original mass of CuO
was much greater than the original mass of NH3.
304 • Chapter 9 • Chemical Quantities
9
Step 4 CuO is the limiting reactant, so we must use the amount of CuO
in calculating the amount of N2 formed. Using the mole ratio
between CuO and N2 from the balanced equation, we have
1 mol N2
1.14 mol CuO ! ___________
" 0.380 mol N2
3 mol CuO
Step 5 Using the molar mass of N2 (28.02), we can now calculate the mass
of N2 produced.
28.02 g N
0.380 mol N2 ! __________2 " 10.6 g N2
1 mol N2
Practice Problem • Exercise 9.8
Lithium nitride, an ionic compound containing the Li! and N3" ions, is
prepared by the reaction of lithium metal and nitrogen gas. Calculate the
mass of lithium nitride formed from 56.0 g of nitrogen gas and 56.0 g of
lithium in the unbalanced reaction:
Li(s) # N2(g) n Li3N(s)
Active Reading Question
In the example, how can CuO be limiting if we started with a greater
mass of CuO than NH3?
C. Percent Yield
Previously, we calculated the amount of product formed when specified
amounts of reactants were mixed together. In doing these calculations, we
used the fact that the amount of product is controlled by the limiting reactant. Products stop forming when one reactant runs out.
The amount of product calculated in this way is called the theoretical
yield of that product. It is the amount of product predicted from the
amounts of reactants used. For instance, in Example 9.8, 10.6 g of nitrogen
represents the theoretical yield. This is the maximum amount of nitrogen
that can be produced from the quantities of reactants used. Actually, however, the amount of product predicted (the theoretical yield) is seldom
obtained. One reason for this is the presence of side reactions (other
reactions that consume one or more of the reactants or products).
The actual yield of product, which is the amount of product actually
obtained, is often compared to the theoretical yield. This comparison,
usually expressed as a percent, is called the percent yield.
Actual yield
________________
! 100% " percent yield
Theoretical yield
For example, if the reaction considered in Example 9.8 actually gave 6.63 g
of nitrogen instead of the predicted 10.6 g, the percent yield of nitrogen
would be
6.63
g N2
_________
! 100% " 62.5%
10.6 g N2
Theoretical yield
The maximum amount of
a given product that can
be formed when the
limiting reactant is
completely consumed
Percent yield
The actual yield of a
product as the percentage
of the theoretical yield
i
nformation
Percent yield is important
as an indicator of the
efficiency of a particular
reaction.
9.3 • Limiting Reactants and Percent Yield • 305
Active Reading Question
Which do we calculate by using stoichiometry, the actual or the
theoretical yield?
EXA MP L E 9. 9
Stoichiometric Calculations: Determining Percent Yield
Methanol can be produced by the reaction between carbon monoxide and
hydrogen. Let’s consider this process again. Suppose 68.5 kg (6.85 ! 104 g)
of CO(g) is reacted with 8.60 kg (8.60 ! 103 g) of H2(g).
MATH
•
Calculate the theoretical yield of methanol.
Part ! 100%
Percent # ______
Whole
•
If 3.57 ! 104 g of CH3OH is actually produced, what is the percent yield
of methanol?
Solution
Where do we want to go?
•
theoretical yield of CH3OH
•
percent yield of CH3OH
What do we know?
•
CO(g) " H2(g) n CH3OH(l)
•
8.60 ! 103 g H2
•
6.85 ! 104 g CO
•
3.57 ! 104 g CH3OH produced
How do we get there?
Step 1 The balanced equation is
2H2(g) " CO(g) n CH3OH(l).
Step 2 Next, we calculate the moles of reactants.
1 mol CO # 2.45 ! 103 mol CO
6.85 ! 104 g CO ! ___________
28.01 g CO
1 mol H2
8.60 ! 103 g H2 ! __________
# 4.27 ! 103 mol H2
2.016 g H2
Step 3 Now we determine which reactant is limiting. Using the mole ratio
between CO and H2 from the balanced equation, we have
2 mol H2
# 4.90 ! 103 mol H2
2.45 ! 103 mol CO ! __________
1 mol CO
Moles of H2
present
4.27 ! 103
306 • Chapter 9 • Chemical Quantities
less
than
Moles of H2
needed to
react with
all the CO
4.90 ! 103
We see that 2.45 ! 103 mol of CO requires 4.90 ! 103 mol of H2.
Because only 4.27 ! 103 mol of H2 is actually present, H2 is limiting.
Step 4 We must therefore use the amount of H2 and the mole ratio between
H2 and CH3OH to determine the maximum amount of methanol that
can be produced in the reaction.
1 mol CH3OH
4.27 ! 103 mol H2 ! ______________
" 2.14 ! 103 mol CH3
2 mol H2
This represents the theoretical yield in moles.
Step 5 Using the molar mass of CH3OH (32.04 g), we can calculate the
theoretical yield in grams.
32.04 g CH3OH
2.14 ! 103 mol CH3OH ! _______________
" 6.86 ! 104 g CH3OH
1 mol CH3OH
So, from the amounts of reactants given, the maximum amount of
CH3OH that can be formed is 6.85 ! 104 g. This is the theoretical yield.
The percent yield is
Actual yield (grams)
________________________
! 100%
Theoretical yield (grams)
3.57 ! 104 g CH3OH
" ____________________
! 100%
6.86 ! 104 g CH3OH
" 52.0%
Practice Problem • Exercise 9.9
Titanium(IV) oxide is a white compound used as a coloring pigment.
In fact, the page you are now reading is white because of the presence
of this compound in the paper. Solid titanium(IV) oxide can be prepared
by reacting gaseous titanium(IV) chloride with oxygen gas. A second
product of this reaction is chlorine gas.
TiCl4(g) # O2(g) n TiO2(s) # Cl2(g)
a. Suppose that 6.71 ! 103 g of titanium(IV) chloride is reacted
with 2.45 ! 103 g of oxygen. Calculate the maximum mass of
titanium(IV) oxide that can form.
b. If the percent yield of TiO2 is 75%, what mass is actually formed?
9.3 • Limiting Reactants and Percent Yield • 307