CP Chemistry
Unit 8 Stoichiometry
Chapter 9
Unit 6b Stoichiometry Standards
3.2.10 B Apply process knowledge and
organize scientific and technological
phenomena in varied ways.
 3.2.10 C Apply the elements of scientific
inquiry to solve problems.
 3.4.10 A Explain concepts about the
structure and properties of matter.

Chapter Themes
The Power of a Balanced Equation
 Stoichiometry
 Limiting Reactants
 Yield (actual / theoretical / percent)
 Day One Assignment
 Day Two … Class Information
 Day Two Assignment

The Power of a Balanced Equation

Discover that a balanced equation provides much
information …
2H2O  2H2 + O2




2 molecules of water = 2 molecules of hydrogen plus 1
molecule of oxygen
2 moles of water = 2 moles of hydrogen plus 1 mole of
oxygen
2 dozen molecules of water = 2 dozen molecules of
hydrogen plus 1 dozen molecules of oxygen
The decomposition of water yields twice as much
hydrogen as oxygen.
Stoichiometry

Definition:
The process of using a chemical equation
to calculate the relative masses of
reactants and products involved in a
reaction.

In ordinary English:
If you know one mass, you can calculate
any other mass!
Stoichiometry


We do this everyday …
Recipe …
Betty Crocker Cake Mix … Double Chocolate
Delight
1 egg
1/3 cup oil
+ 8 X 12 greased cake pan
1 Double Chocolate Delight Cake
1 egg + 1/3 cup oil  1 D.C.D.C.
Stoichiometry
We do this everyday …
 Recipe …
 Milage …

11 gallons gas
+ 1 ’97 Neon
382 miles driven
11 gallons gas + 1 Neon  382 miles
Limiting Reactants

Consider sandwiches …


1 sandwich =
2 pieces of bread
1 piece cheese
3 slices of meat
How many sandwiches can you make
with 7 pieces of bread; 5 pieces of
cheese; and 25 slices of meat?
Yield (actual / theoretical / percent)

Back to the ’97 Neon …
- Do I always get 382 miles / 11 gallons?

What do you think these words mean?



Actual yield …
Theoretical yield …
Percent yield …
Day One Assignment
p. 282: 3a&d; 4b&d; 10b&d
 3a
2NO + O2  2NO2
2 molecules NO + 1 molecule O2 yields 2 molecules NO2
2 moles NO + 1 mole O2 yields 2 moles NO2

3d
C2H6 + Cl2  C2H5Cl + HCl
1 molecule C2H6 + 1 molecule Cl2 yields 1 molecule C2H5Cl +
1 molecule HCl
1 mole C2H6 + 1 mole Cl2 yields 1 mole C2H5Cl + 1 mole HCl
Day One Assignment
p. 282: 3a&d; 4b&d; 10b&d
 4b B2O3 + 3CaF2  2BF3 + 3CaO
1 molecule B2O3 + 3 molecule CaF2 yields 2 molecules BF3
+ 3 molecules CaO
1 mole B2O3 + 3 moles CaF2 yields 2 moles BF3 + 3 moles
CaO

4d C6H6 + 3H2  C6H12
1 molecule C6H6 + 3 molecules hydrogen gas yields 1
molecule C6H12
1 mole C6H6 + 3 moles hydrogen gas yields 1 mole C6H12
Day One Assignment
p. 282: 3a&d; 4b&d; 10b&d
 10b SeO2 + 2H2Se  3Se + 2H2O
0.625 mole H2O X 3 mole Se/2 mole H2O = 0.938 mole Se

10d Fe2O3 + 2Al  2Fe + Al2O3
0.625 mole Al2O3 X 2 moles Fe/1 mole Al2O3 = 1.25 mole
Fe
Day Two … Class Information

The Meaning of a Balanced Chemical Equation
2H2O  2H2 + O2




2 molecules of water = 2 molecules of hydrogen plus 1
molecule of oxygen
2 moles of water = 2 moles of hydrogen plus 1 mole of
oxygen
2 dozen molecules of water = 2 dozen molecules of
hydrogen plus 1 dozen molecules of oxygen
The decomposition of water yields twice as much
hydrogen as oxygen … MOLE RATIOs
Day Two … Class Information
2H2O  2H2 + O2
2 moles of H2O = 2 moles H2 + 1 mole O2
6 moles of H2O = __ moles H2 + __ moles
O2
 6 moles of H2O = 6 moles H2 + 3 moles O2

½ mole H2O = __ mole H2 + __ mole O2
 ½ mole H2O = ½ mole H2 + ¼ mole O2

Day Two … Class Information
2NH3  N2 + 3H2
2 moles NH3  1 moles N2 + 3 moles H2
4 moles NH3  __ moles N2 + __ moles H2
4 moles NH3 (1 mole N2 / 2 moles NH3) = 2 moles N2
4 moles NH3 (3 mole H / 2 moles NH3) = 6 mole H2
Day Two … Class Information
3MnO2 + 4Al  3Mn + 2Al2O3
3 moles MnO2 + 4 moles Al  3 moles Mn +
2 moles Al2O3
0.685 moles Al = ______ moles Mn
0.685 moles Al (3 moles Mn / 4 moles Al) =
0.514 moles Mn
Day Two Assignment
p. 282: 8b&d; 13d&g; 14b&d
 8b Cl2 + 2KI  2KCl + I2
0.125 mole KI X 2 mole KCl/2 mole KI = 0.125 mole KCl
0.125 mole KI X 1 mole I2 /2 mole KI = 0.0625 mole I2

8d CaC2 + 2H2O  Ca(OH)2 + C2H2
0.125 mole CaC2 X 1 mole Ca(OH)2/1 mole CaC2 = 0.125
mole Ca(OH)2
0.125 mole CaC2 X 1 mole C2H2 / 1 mole CaC2 = 0.125
mole C2H2
Day Two Assignment
p. 282: 8b&d; 13d&g; 14b&d
 13d molar mass of SO2 = 64.07 g/mole
(5.23 kg SO2)(1000g/1 kg)(1 mole/64.07g) = 81.6 mole
SO2

13g molar mass of LiOH = 23.95 g/mole
(6.91 X 103 g LiOH)(1 mole/23.95 g) = 288.5 = 289 mole
LiOH
Day Two Assignment
p. 282: 8b&d; 13d&g; 14b&d
 14b molar mass of He = 4.003 g/mole
(2.75 mole He)(4.003 g/1 mole He) = 11.0 g He

14d molar mass of CO2 = 44.01 g/mole
(7.21 X 10-3 mole CO2)(44.01 g/1 mole CO2) = 0.317 g
CO2