1. Stoichiometry
Chapter 11

2. Chemical Reactions/Equations
REVIEW
Chemical Reactions/Equations
Reactant + Reactant → Product + Product
Balanced chemical equations are written to show chemical changes taking place
Balanced chemical equations reflect the Law of Conservation of Mass
Coefficients balance chemical equations and indicate the relative amounts of substances

3. Conversion Factors REVIEW Avogadro’s number and moles Molar mass
1 mol = 6.02 x 1023 representative particles
Representative particles:
Atoms – element from periodic table
Ions – charged atom (due to loss/gain of e–)
Formula units – ionic compound
Molecules – covalent compound
Molar mass
1 mol =  g of substance
where  = (atomic mass × subscript)
Mole ratio of element to compound
 mol element = 1 mol compound
where  = subscript of element in compound

4. Unit Objectives PREVIEW
Interpret chemical equations in terms of particles, moles, and mass
Write mole ratios from balanced equations
Calculate number of moles and mass of a reactant or product when given number of moles or mass of another reactant or product
Identify limiting reactants in chemical reactions
Determine percent yield of chemical reactions

5. Stoichiometry U8-5 Defined: Based on the Law of Conservation of Mass:
The study of quantitative relationships between amounts of reactants used and products formed by a chemical reaction
Involves calculating quantities of reactants or products in a reaction using relationships found in balanced chemical equation
Based on the Law of Conservation of Mass:
Mass is neither created nor destroyed in any process; it is conserved

6. Interpreting Equations
Write the balanced chemical equation for the synthesis of ammonia from nitrogen and hydrogen gases.
Interpret the equation in terms of  particles,  moles, and  mass.
Then,  show the Law of Conservation of Mass is obeyed. N2 H2 3 →
NH3 2

7. Interpreting Equations
 Interpret the equation in terms of particles.
Number of particles for each substance is indicated by the coefficients
Four particle types: atoms, ions, formula units, or molecules
N H2 → 2 NH3
1 molecule N molecules H2 → 2 molecules NH3

8. Interpreting Equations
 Interpret the equation in terms of moles.
Number of moles for each substance is indicated by the coefficients
N H2 → 2 NH3
1 mole N moles H2 → 2 moles NH3

9. Interpreting Equations
 Interpret the equation in terms of mass.
Calculate the molar mass for each reactant and product
N2:
H2:
NH3:
N H2 → 2 NH3
2 mol x 14.0 g/mol
28.0 g/mol N2
2 mol x g/mol
6.0 g/mol H2
1 mol x 14.0 g/mol = 14.0 g
3 mol x g/mol = g
17.0 g/mol NH3

10. Interpreting Equations
 Interpret the equation in terms of mass.
Multiply number of moles of each reactant and product by molar mass
N H2 → 2 NH3
28.0
g/mol
2.0
g/mol
17.0
g/mol
1 mol N2
28.0 g N2
g N2 =
= 28.0 g N2
1 mol N2

11. Interpreting Equations
 Interpret the equation in terms of mass.
Multiply number of moles of each reactant and product by molar mass
N H2 → 2 NH3
28.0
g/mol
2.0
g/mol
17.0
g/mol
3 mol H2
2.0 g H2
g H2 =
= 6.0 g H2
1 mol H2

12. Interpreting Equations
 Interpret the equation in terms of mass.
Multiply number of moles of each reactant and product by molar mass
N H2 → 2 NH3
28.0
g/mol
2.0
g/mol
17.0
g/mol
2 mol NH3
17.0 g NH3
g NH3 =
= 34.0 g NH3
1 mol NH3

13. Interpreting Equations
 Interpret the equation in terms of mass.
N H2 → 2 NH3
28.0 g
6.0 g
34.0 g
28.0 g N g H2 → g NH3

14. Interpreting Equations
 Show that the Law of Conservation of Mass is observed.
Add the masses of the reactants.
28.0 g N g H2 = g reactants
Add the masses of the products.
34.0 g NH3 = g products
When the mass of the reactants equals the mass of the products, the Law of Conservation of Mass is observed.
34.0 g reactants = g products
N H2 → 2 NH3
28.0 g
6.0 g
34.0 g

15. HW due 01/28
Textbook Practice, p. 371
#1, #2

16. U8-5
Mole Ratio
Ratio between the numbers of moles of any two substances in a balanced chemical equation

17. U8-5
Mole Ratios
Determine all possible mole ratios for the balanced equation showing the synthesis of ammonia.
How many mole ratios are possible?
This reaction has three participating species.
Multiply the number of species present by the next lower number.
For the synthesis of ammonia, six mole ratios are possible.
N H2 → 2 NH3
3 x 2 = 6 mole ratios

18. U8-5
Mole Ratios
Determine all possible mole ratios for the balanced equation showing the synthesis of ammonia.
N H2 → 2 NH3
1 mol N2
3 mol H2
3 mol H2
2 mol NH3
2 mol NH3
1 mol N2
1 mol N2
2 mol NH3
3 mol H2
1 mol N2
2 mol NH3
3 mol H2

19. Mole Ratios U8-5 GIVEN Which mole ratio should be used?
The needed mole ratio is the one involving the UNKNOWN and the GIVEN.
UNKNOWN
GIVEN

20. HW due 01/29
Textbook Practice, p. 372 #3

21. Stoichiometric Calculations
Write the balanced chemical equation for the reaction.
Identify the UNKNOWN and the GIVEN; draw the bridge/grid.
The GIVEN must be in moles or converted to moles.
Identify the conversion factor that will cancel the unit of the GIVEN.
Set up conversion factors and cancel units until the only unit left standing matches the UNKNOWN.
Do the math* and express the answer to the correct number of significant figures.

22. Mole-to-Mole Conversions
Both the UNKNOWN and the GIVEN are in moles
Example:
How many moles of ammonia are produced when 10.0 moles of hydrogen react with excess nitrogen?
10.0 mol
? mol
excess
N H2 → 2 NH3

23. Mole-to-Mole N2 + 3 H2 → 2 NH3 10.0 mol ? mol
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
10.0 mol H2
mol NH3 2
mol NH3 =
mol H2 3
= 6.67 mol NH3

24. Mole-to-Mole Conversions Practice 1
How many moles of zinc chloride will be formed when 17.0 moles of hydrochloric acid react with excess zinc metal? How many mole ratios are possible for this equation?
17.0 mol
? mol
excess 2
HCl
+ Zn →
ZnCl2 H2 12

25. Mole-to-Mole Conversions Practice 1
2 HCl + Zn → ZnCl2 + H2
Write all possible mole ratios for the balanced equation.
2 mol HCl
1 mol Zn
1 mol ZnCl2
1 mol H2

26. Mole-to-Mole 2 HCl + Zn → ZnCl2 + H2 17.0 mol ? mol
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
17.0 mol HCl
1 mol ZnCl2
mol ZnCl2 =
2 mol HCl
= 8.50 mol ZnCl2

27. Mole-to-Mole Conversions Practice 2
Potassium chlorate decomposes into potassium chloride and oxygen. How many moles of oxygen are formed when 3.20 moles KClO3 decompose? Write the mole ratios involving the UNKNOWN and the GIVEN in the problem.
3.20 mol
? mol 2
KClO3 →
KCl 2 O2 3
3 mol O2
2 mol KClO3
2 mol KClO3
3 mol O2

28. Mole-to-Mole 2 KClO3 → 2 KCl + 3 O2 3.20 mol ? mol
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
3.20 mol KClO3
3 mol O2
mol O2 =
2 mol KClO3
= 4.80 mol O2

29. HW due 02/04
Textbook Practice, p. 375
#11

30. Mole-to-Mass Conversions
p. 4 of Stoichiometry for Students Notes Packet

31. Mole-to-Mass Conversions
GIVEN in moles and UNKNOWN in grams
Example: Balance the following equation for the combustion of propane.
Calculate the molar mass for each substance. 3 4 5
____ C3H8 + ____ O2 → ____ CO2 + ____ H2O
44.0 g/mol
32.0 g/mol
44.0 g/mol
18.0 g/mol

32. Mole-to-Mass Conversions
Example: If 10.0 moles of propane are used, how many grams of water are formed? Write the mole ratios involving the UNKNOWN and the GIVEN.
10.0 mol
? g
C3H O2 → 3 CO H2O
1 mol C3H8
4 mol H2O
4 mol H2O
1 mol C3H8

33. Mole-to-Mass C3H8 + 5 O2 → 3 CO2 + 4 H2O 10.0 mol ? g
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
10.0 mol C3H8
4 mol H2O
18.0 g H2O
g H2O =
1 mol C3H8
1 mol H2O
= 720. g H2O

34. Mole-to-Mass Conversions
Practice 1: Sulfuric acid is produced when sulfur dioxide reacts with oxygen and water. Balance the equation for the synthesis reaction.
Calculate the molar mass for each substance.
____ SO2 + ____ O2 + ____ H2O → ____ H2SO4 2
64.1 g/mol
32.0 g/mol
18.0 g/mol
98.1 g/mol

35. Mole-to-Mass Conversions
Practice 1: How many grams of sulfuric acid are produced when 1.50 moles of sulfur dioxide completely reacts with oxygen and water ? Write the mole ratios involving the UNKNOWN and the GIVEN.
1.50 mol
? g
2 SO2 + O H2O → 2 H2SO4
2 mol H2SO4
2 mol SO2
2 mol SO2
2 mol H2SO4

36. Mole-to-Mass 2 SO2 + O2 + 2 H2O → 2 H2SO4 ? g 1.50 mol
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
1.50 mol SO2
2 mol H2SO4
98.1 g H2SO4
g H2SO4 =
2 mol SO2
1 mol H2SO4
= 147 g H2SO4

37. Mass-to-Mole Conversions
GIVEN in grams and UNKNOWN in moles
Example: Methane and sulfur produce carbon disulfide and hydrogen sulfide gas, as indicated by the following chemical equation.
Calculate the molar mass for each substance.
____ CH4 + ____ S8 → ____ CS2 + ____ H2S 2 4
16.0 g/mol
256.8 g/mol
76.2 g/mol
34.1 g/mol

38. Mass-to-Mole Conversions
Example: Suppose that g sulfur react with an excess of methane. How many moles of carbon disulfide will be produced?
19.75 g
? mol
2 CH4 + S8 → 2 CS H2S

39. Mass-to-Mole 2 CH4 + S8 → 2 CS2 + 4 H2S ? mol 19.75 g
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
19.75 g S8
1 mol S8
2 mol CS2
mol CS2 =
256.8 g S8
1 mol S8
= mol CS2

40. Mass-to-Mole Conversions
Practice 1: Sodium fluoride is formed when sodium metal reacts with fluorine gas.
Calculate the molar mass for each substance.
Balance the equation.
____ Na + ____ F2 → ____ NaF 2
23.0 g/mol
38.0 g/mol
42.0 g/mol

41. Mass-to-Mole Conversions
Practice 1: How many moles of sodium fluoride can be formed when 4.57 grams of fluorine gas reacts completely with excess sodium?
4.57 g
? mol
2 Na + F2 → 2 NaF

42. Mass-to-Mole 2 Na + F2 → 2 NaF ? mol 4.57 g
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
4.57 g F2
1 mol F2
2 mol NaF
mol NaF =
38.0 g F2
1 mol F2
= mol NaF

43. HW due 02/05
Textbook Practice, p. 376
#14

44. Mass-to-Mass Conversions
GIVEN in grams and UNKNOWN in grams
Example: The following equation shows the combustion of butane.
Calculate the molar mass for each substance.
Balance the equation.
____ C4H ____ O2 → ____ CO2 + ____ H2O 8 10 2 13
58.0 g/mol
32.0 g/mol
44.0 g/mol
18.0 g/mol

45. Mass-to-Mass Conversions
Example: If 75.5 grams of carbon dioxide are produced, how many grams of butane were used?
? g
75.5 g
2 C4H O2 → 8 CO H2O

46. Mass-to-Mass 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O ? g 75.5 g
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
75.5 g CO2
1 mol CO2
2 mol C4H10
58.0 g C4H10
g C4H10 =
44.0 g CO2
8 mol CO2
1 mol C4H10
= 24.9 g C4H10

47. Mass-to-Mass Conversions
Practice 1: Use the balanced equation for the combustion of butane. How many grams of oxygen are necessary to react completely with 217 grams of butane?
217 g
? g
2 C4H O2 → 8 CO H2O

48. Mass-to-Mass 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O 217 g ? g
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
217 g C4H10
1 mol C4H10
13 mol O2
32.0 g O2
g O2 =
58.0 g C4H10
2 mol C4H10
1 mol O2
= 778 g O2

49. Mass-to-Mass Conversions
Practice 2: Balance the following single replacement reaction.
How many grams of iron must react in order to produce 75.9 grams of iron (III) oxide?
Because grams are involved for both UNKNOWN and GIVEN, the molar mass for both must be calculated.
? g
75.9 g
____ Fe + ____ H2O → ____ Fe2O3 + ____ H2 3 2
Molar mass: UNKNOWN
Molar mass: GIVEN
55.8 g/mol
159.6 g/mol

50. Mass-to-Mass 2 Fe + 3 H2O → Fe2O3 + 3 H2 ? g 75.9 g
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
75.9 g Fe2O3
1 mol Fe2O3
2 mol Fe
55.8 g Fe
g Fe =
159.6 g Fe2O3
1 mol Fe2O3
1 mol Fe
= 53.1 g Fe

51. HW due 02/06
Textbook Practice, p. 378
#21