Monday, November 26, 2007 -- Bellringer
You are building model cars. Each car has four
tires, two doors, and one body. If you have 16
tires, 10 doors, and 7 car bodies, how many model
cars can you build? What is left over after you
have built all of the cars?
Answer: You can build 4 cars and will have 2 doors and
3 bodies left over
Remember the mole . . .
Known
Unknown
Substance A
Substance B
Mass
Mass
Mole
Mole
Particles
Particles
Remember the mole . . .
Known
Unknown
Substance A
Substance B
Mass
Mass
Use coefficients
from balanced
chemical equation
Mole
Particles
Mole
Particles
Section 9.1 – Using Chemical Equations
Because atoms are rearranged (not created or destroyed)
in a chemical reaction, we must always balance a chemical
equation.
2 H2(g) + O2(g)  2 H2O (l)
This equation tells us that for every oxygen molecule, two
hydrogen molecules are required to form two molecules of
water.
2 H2(g) + O2(g)  2 H2O (l)
It is important to recognize that the coefficients in a balanced
equation give the relative numbers of molecules
We can multiply each coefficient by any number and the
equation remains balanced. For example, if we multiply by 12:
24 H2(g) + 12 O2(g)  24 H2O (l)
What does a chemical equation tell us:
2 H2(g) + O2(g)  2 H2O (l)
2 H2(g)

O2(g)
2 H2O (l)
2 molecules H2
1 molecule O2

2 molecules H2O
2 dozen molecules H2
1 dozen molecules O2

2 dozen molecules H2O
2 x 6.02x1023 molecules H2 6.02x1023 molecules O2 
2 moles H2
1 moles O2

2 x 6.02x1023 molecules H2O
2 moles H2O
Mole-Mole Relationships
Mole Ratio – The ratio of moles of one substance
to moles of another substance in a balanced
chemical equation.
Back to our car example. Write an equation to
represent the parts needed to assemble car:
Four tires, two doors, one body for each car
4T + 2D + B  T4D2B
12T + 6D + 3B  _________T4D2B
How many moles of ammonia can be produced
with 4.3 moles of nitrogen?
N2(g) + H2(g)  NH3(g)
Balance the equation:
N2(g) + 3 H2(g)  2 NH3(g)
How many moles of nitrogen gas are there
compared to moles of ammonia in the balanced
equation?
4.3 mol N 2 x
2 mol NH3
 ______ mol NH3
1 mol N 2
Practice Problem: What number of moles of O2(g) is
required to react with 3.6 mole SO2(g) in the following
reaction?
O2(g) + 2 SO2(g)  2 SO3(g)
___
1 1 mol
mol
molO
OO
22 2
3.6
3.6mol
molSO
SO
______
______
1.8 _ mol
mol
mol
OO
2 x
2 x
2O
22
___
2 2 mol
mol
molSO
SO
SO
2 22
Homework: Chapter 9 Assessment (p. 310, #2,3,6,8,9,10)
Section 9.2 – Using Chemical Equations to Calculate Mass
Problem: Calculate the mass of iodine needed to just
react with 35.0 g of aluminum.
Al(s) + I2(s)  AlI3(s)
What do we know?
Al(s) + I2(s)  AlI3(s)
35.0 g aluminum
How do we solve this problem?
Balance the equation:
2 Al(s) + 3 I2(s)  2 AlI3(s)
What next?
Map it out!
35.0 g Al
? grams I2
? Moles Al
? Moles I2
Solve it!
•Convert grams of aluminum to moles of aluminum
___1__mol
______mol Al
Al
35
 __
1.30 mol
35 gg Al
Al xx
______
molAl
Al
__26.8_g Al
_______g
Al
•Determine the number of moles of iodine using the
coefficients in the balanced chemical equation:
____
__3_ mol I 2
11..30
30mol
mol Al
Al xx
_____
_ 1.95 mol I 2
__2__
____ mol
molAl
Al
•Use the molar mass of iodine to calculate the grams of
iodine needed:
___
g gI 2 I 2
253.8
1.195
 ______g
.95mol
molI 2I 2 xx
495 g I 2I 2
__1mol
mol II22
Steps to calculate the mass of reactants
and products in a chemical reaction
1. Balance the equation for the reaction.
2. Convert the masses of reactants or products to
moles
3. Use the balanced equation to set up the appropriate
mole ratio(s)
4. Use the mole ratio(s) to calculate the number of
moles of the desired reactant or product.
5. Convert from moles back to mass.
Stoichiometry – The process of using a balanced chemical
equation to determine the relative masses of reactants and
products involved in a reaction.
Homework: Chapter 9 Assessment (p. 310, #11, 13a-d, 14a-d, 15a-b, 16a-b, 17, 18, 21, 24)
Section 9.3 – Limiting Reactants and Percent Yield
More model cars:
4 tires + 2 doors + 1 body = 1 car
How many cars can be made if you had:
40 tires
16 doors
12 bodies
Determine how many cars can be made with each of the parts:
11car
car
 ______
__ 10 _ cars
44 tires
tires
11car
car
16
 __
8 ___ cars
cars
16doors
doors xx
______
22doors
doors
40
40 tires
tires xx
12 bodies x
1 car
 ___
______
12 _cars
cars
1 body
How many cars can be made?
8 cars
What is left over?
8 tires
4 bodies
N2(g)
+
3H2(g) 
2NH3(g)
Container (#1) of N2(g) and H2(g).
Before and after the reaction.
N2(g)
+
3H2(g) 
2NH3(g)
Container (#2) of N2(g) and H2(g).
Before and after the reaction.
Limiting Reactant
Limiting Reactant – The reactant that is completely used
up when a reaction is run to completion.
Steps for solving stoichiometry problems involving
limiting reactants
1. Write and balance the equation for the reaction.
2. Convert known masses of reactants to moles
3. Using the numbers of moles of reactants and the appropriate
mole ratios, determine which reactant is limiting.
4. Using the amounts of the limiting reactant and the
appropriate mole ratios, compute the number of moles of
the desired product.
5. Convert from moles of product to grams of product, using
the molar mass.
Sample Calculation
2Cu(s) + S(s)  Cu2S(s)
80g
25g ? G
Solve for mol Cu2S using both reagents and
compare. Choose the smaller.
1 mol Cu 1 mol Cu 2S 159.1 g
80.0g Cu x
x
x
 100. g Cu 2S
63.5g Cu 2 mol Cu
1 mol
1 mol S 1 mol Cu 2S 159.1 g
25.0 g S x
x
x
 124 g Cu 2S
32.1 g S
1 mol S
1 mol
Percent Yield
Theoretical Yield – The maximum amount of a given
product that can be formed when the limiting
reactants is completely consumed.
Percent Yield – The actual yield of a product as the
percentage of the theoretical yield.
Actual Yield
x 100  Percent Yield
Theoretica l Yield