1. Chemical Reactions

2. Chemical Equations 4 Al (s) + 3 O2 (g)  2 Al2O3 (s)
A chemical reaction shows the formulas and relative amounts of reactants and products in a reaction.
4 Al (s) O2 (g)  2 Al2O3 (s)
reactants
product
stoichiometric
coefficients
physical state
s, l, g or aq

3. Chemical Equations 4 Al (s) + 3 O2 (g)  2 Al2O3 (s)
Interpret this equation as
4 atoms solid Al react with 3 molecules gaseous O2 to form 2 formula units of solid Al2O3
4 moles solid Al react with 3 moles gaseous O2 to form 2 moles of solid Al2O3

4. Balancing Chemical Equations
“Matter is conserved in
chemical change”
Antoine Lavoisier, 1789
An equation must be balanced:
It must have the same number
of atoms of each kind on both sides

5. The rules of the game
Write the correct formulas of the reactants and products
Do not change the formulas to balance the equation
Put a coefficient in front of each formula so that the same number of atoms of each kind appear in both the reactants and products
The coefficent multiplies through its formula
2 H2O shows 4 H atoms and 2 O atoms

6. Example 1 Balance this equation: N2 + H2  NH3 N2 + H2  2 NH3
There are 2 Ns on the left, we need 2 Ns on the right
Put a coefficient of 2 in front of NH3
N2 + H2  2 NH3
Now there are 6 Hs on the right, we need 6 on the left
Put a coefficient of 3 in front of H2
N2 + 3 H2  2 NH3
Balanced equation shows 2 Ns and 6 Hs on each side.

7. Example 2 Balance this equation: CH4 + O2  CO2 + H2O
There is one C on each side. Cs are balanced
There are 4 Hs on left, 2 on right. Put a 2 in front of H2O
CH4 + O2  CO2 + 2 H2O
There are = 4 Os on right. Put a 2 in front of O2
CH4 + 2 O2  CO2 + 2 H2O
Balanced equation has 1 C, 4 Hs, and 4 Os on each side

8. Mg (s) + HCl (aq)  MgCl2 (aq) + H2 (g)
Balance these equations and list the coefficients from left to right (including coefficients of one)
HgO (s)  Hg (l) + O2 (g)
2 HgO (s)  2 Hg (l) + O2 (g) Answer 2, 2, 1
Mg (s) + HCl (aq)  MgCl2 (aq) + H2 (g)
Mg (s) HCl (aq)  MgCl2 (aq) + H2 (g) Answer 1, 2, 1, 1

9. C3H8 (g) + O2 (g)  CO2 (g) + H2O (l)
Balance these equations and list the coefficients from left to right (including coefficients of one)
C3H8 (g) + O2 (g)  CO2 (g) + H2O (l)
C3H8 (g) O2 (g)  3 CO2 (g) H2O (l) Answer 1, 5, 3, 4
Al (s) + Cl2 (g)  AlCl3 (s)
2 Al (s) Cl2 (g)  2 AlCl3 (s) Answer 2, 3, 2

10. C4H10 (g) + O2 (g)  CO2 (g) + H2O (l)
Balance these equations and list the coefficients from left to right (including coefficients of one)
H2O2 (aq)  H2O (l) + O2 (g)
2 H2O2 (aq)   H2O (l) + O2 (g) Answer 2, 2, 1
C4H10 (g) + O2 (g)  CO2 (g) + H2O (l)
2 C4H10 (g) O2 (g)  8 CO2 (g) H2O (l) Answer 2, 13, 8, 10

11. Combustion Reactions
In combustion, a hydrocarbon or C–H–O fuel combines with O2 to form CO2 and H2O
__ CH4 + __ O2  __ CO2 + __ H2O
1 CH O2  1 CO H2O
Balanced equation shows 1 C, 4 H, and 4 O on each side
If N or S are in the formula for the fuel, assume it is oxidized to NO2 or SO2

12. Reaction types Synthesis or combination A + B → C
2 or more formulas combine to make one product
Decomposition C → A + B
One formula breaks apart into 2 or more formulas
Combustion
C, H, O fuel + O2 → CO2 + H2O
Single replacement A + BX → AX + B
One element pushes another element out of a compound
Double replacement AX + BY → AY + BX
Two compounds switch partners

13. Reaction types
Classify each reaction as combination, decomposition, combustion, single replacement, or double replacement:
2 C2H O2 → 4 CO H2O
2 Na + Cl2 → 2 NaCl
Cl KBr → 2 KCl + Br2
NH4NO3 → N2O H2O
K2SO4 + BaCl2 → BaSO KCl

14. Stoichiometry Stoichiometry is chemical accounting
Heart of stoichiometry is mole ratio given by coefficients of balanced equation

15. Stoichiometry Stoichiometry is chemical accounting
Heart of stoichiometry is mole ratio given by coefficients of balanced equation
moles A
moles B
mole ratio

16. Example
How many moles of O2 are produced from the decomposition of 1.76 mol KClO3: 2 KClO3 (s)  2 KCl (s) + 3 O2 (g)
Question about O2 but information for KClO3
O2 and KClO3 related by coefficients of balanced equation.
Coefficients are in moles: 2 mol KClO3 : 3 mol O2

17. Example
How many moles of O2 are produced from the decomposition of 1.76 mol KClO3: 2 KClO3 (s)  2 KCl (s) + 3 O2 (g)

18. Example
How many moles of CO2 form when 0.35 mol C2H6 burn? C2H O2 → 4 CO H2O

19. Example
How many moles of CO2 form when 0.35 mol C2H6 burn? C2H O2 → 4 CO H2O

20. Example
How many grams of Al react with 0.15 moles HCl: 2 Al (s) HCl (aq)  2 AlCl3 (aq) H2 (g)
Coefficients are in moles
Two steps: convert mol HCl to mol Al, then convert mol Al to grams

21. Example
How many grams of Al react with 0.15 moles HCl: 2 Al (s) HCl (aq)  2 AlCl3 (aq) H2 (g)
Coefficients are in moles
Two steps: convert mol HCl to mol Al, then convert mol Al to grams

22. Example
How many moles of Ag are produced when 10.0 g Ag2O decompose: Ag2O (s)  4 Ag (s) + O2 (g)
Two steps: first convert g Ag2O to mol Ag2O, then use mole ratio to find moles Ag

23. Example
How many moles of Ag are produced when 10.0 g Ag2O decompose: Ag2O (s)  4 Ag (s) + O2 (g)
Two steps: first convert g Ag2O to mol Ag2O, then use mole ratio to find moles Ag

24. Example
How many grams of magnesium nitride are produced when 3.82 g Mg react with excess N2: 3 Mg (s) + N2 (g)  Mg3N2 (s)

25. Example
How many grams of H2 (g) are needed to produce g of CH3OH: CO (g) H2 (g)  CH3OH (l)

26. Example
How many grams of H2 react with 3.20 grams of O2 in the reaction 2 H2 (g) + O2 (g)  2 H2O (l)

27. Example
How many grams of O2 react with 268 grams of octane (C8H18) in the combustion of octane? 2 C8H O2  16 CO H2O

28. Example
How many grams of O2 react with 268 grams of octane (C8H18) in the combustion of octane? 2 C8H O2  16 CO H2O

29. Stoichiometry Convert to moles (÷ molar mass) Use the mole ratio
Convert new moles back to grams (x new molar mass)
grams A
moles A
moles B
grams B
mole ratio

30. Example Balance the chemical equation Identify the type of reaction
Calculate how many moles of AgNO3 are needed to produce 3.17 g Ag2CO AgNO3 + K2CO3  Ag2CO3 + KNO3

31. Example Balance the chemical equation Identify the type of reaction
Calculate how many grams of Al are needed to produce g H Al (s) + HCl (aq)  AlCl3 (aq) + H2 (g)

32. 2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g)
Example 4-6B
The model problem describes an Al-Cu alloy composed of 93.7% Al and 6.3% Cu by mass, with a density of 2.85 g/cm3. The Al (but not the Cu) reacts with HCl:
2 Al (s) HCl (aq)  2 AlCl3 (aq) H2 (g)
How many grams of Cu are present in a sample of alloy that yields 1.31 g H2 when it reacts with HCl?

33. Example 4-7A
The model problem describes an HCl solution which is 28% HCl by mass and has a density of 1.14 g/mL. It reacts with Al: Al (s) HCl (aq)  2 AlCl3 (aq) H2 (g)
How many mg of H2 are produced when 1 drop (0.05 mL) of the HCl solution reacts with Al?

34. Example 4-7B
A vinegar contains 4.0% HC2H3O2 by mass and has a density of 1.01 g/mL. It reacts with sodium hydrogen carbonate: HC2H3O2 (aq) + NaHCO3 (s)  NaC2H3O2 (aq) + H2O (l) + CO2 (g)
How many grams of CO2 are produced by the reaction of 5.00 mL of this vinegar with NaHCO3?

35. Chemical Reactions in Solution
Most reactions occur in aqueous solution
SOLUTE is the substance to be dissolved in solution
SOLVENT is the substance (often a liquid) the solute dissolves in
The concentration of the solution is Molarity (M) = moles solute L solution

36. Example 4-8A
If 22.3 g acetone, (CH3)2CO, are dissolved in enough water to make 1.25 L of solution, what is the concentration (M) of the solution?

37. Example 4-8B
15.0 mL of concentrated acetic acid, HC2H3O2 (d = g/mL), are dissolved in enough water to produce mL of solution. What is the concentration of the solution?

38. Example 4-9A
At 25 °C, an aqueous solution saturated with NaNO3 is 10.8 M NaNO3. How many grams of NaNO3 are present in 125 mL of this solution?

39. Example 4-9B
How many grams of Na2SO4 • 10 H2O are needed to prepare 355 mL of M Na2SO4?

40. Dilution problems
It is common to prepare a solution by diluting a more concentrated solution (the stock solution).
The moles of solute taken from the stock solution are given by moles solute = volume x molarity
All the solute taken from the stock appears in the diluted solution, so moles solute are constant: VstockMstock = VdiluteMdilute

41. Example 4-10A
15.00 mL of M K2CrO4 solution are diluted to mL. What is the concentration of the dilute solution?

42. Example 4-10B
After being left out in an open beaker, 275 mL of M NaCl has evaporated to only 237 mL. What is the concentration of the solution after evaporation?

43. Stoichiometry in Solution
Stoichiometry in solution is just the same as for mass problems, except the conversion into or out of moles uses molarity instead of molar mass:
grams A
mL A
moles A
moles B
grams B
mL B
mole ratio

44. Example 4-11A
K2CrO4 (aq) + 2 AgNO3 (aq)  Ag2CrO4 (s) + 2 KNO3 (aq)
How many mL of M K2CrO4 must react with excess AgNO3 to produce 1.50 g Ag2CrO4?

45. Example 4-11B
K2CrO4 (aq) + 2 AgNO3 (aq)  Ag2CrO4 (s) + 2 KNO3 (aq)
How many mL of M AgNO3 must react with excess K2CrO4 to produce exactly 1.00 g Ag2CrO4?

46. Limiting reactant
In a given reaction, often there is not enough of one reactant to use up the other reactant completely
The reactant in short supply LIMITS the quantity of product that can be formed

48. Goldilocks Chemistry
Imagine reacting different amounts of Zn with mol HCl: Zn (s) HCl (aq)  ZnCl2 (aq) + H2 (g)
Rxn 1 Rxn 2 Rxn 3
Mass Zn g g g
Moles Zn mol mol mol
Moles HCl mol mol mol
Ratio mol HCl
mol Zn

49. Limiting reactant problems
The easiest way to do these is to do two stoichiometry calculations
Find the amount of product possible from each reactant
The smaller answer is the amount of product you can actually make (you just ran out of one reactant)
The reactant on which that answer was based is the limiting reactant

50. Example 4-12A
How many grams of PCl3 form when 215 g P4 react with 725 g Cl2: P4 (s) Cl2 (g)  4 PCl3 (l)

51. Example 4-12B
How many kg of POCl3 form if 1.00 kg of each reactant are allowed to react: PCl3 (l) + 6 Cl2 (g) + P4O10 (s)  10 POCl3 (l)

52. Example 4-13A When 215 g P4 react with 725 g Cl2
P4 (s) Cl2 (g)  4 PCl3 (l) (example 4-12A)
which reactant is in excess and what mass of that reactant remains after the reaction is finished?

53. Example 4-13B
12.2 g H2 and 154 g O2 are allowed to react. Identify the limiting reactant, which gas remains after the reaction, and what mass of it is left over H2 (g) + O2 (g)  2 H2O (l)

54. Percent Yield
In real experiments we often do not get the amount of product we calculate we should, because
the reactants may participate in other reactions (side reactions) that produce other products (by-products)
The reaction often does not go to completion.
Percent yield tells the ratio of actual to theoretical amount formed.

55. Percent Yield
Suppose you calculate that a reaction will produce 50.0 g of product. This is the theoretical yield.
The reaction actually produces only 45.0 g of product . This is the actual yield.
Percent yield = g (actual) x 100 = 90.0% g (theoretical)

56. Example 4-14A
If 25.7 g CH2O is produced per mole CH3OH that reacts, what are the theoretical, actual, and percent yield: CH3OH (g)  CH2O (g) + H2 (g)

57. Example 4-14B
What is the percent yield if 25.0 g P4 reacts with 91.5 g Cl2 to produce 104 g PCl3: P4 (s) Cl2 (g)  4 PCl3 (l)

58. Example 4-15A
If the observed percent yield for the formation of urea is 87.5%, what mass of CO2 must react with excess NH3 to produce 50.0 g CO(NH2)2: NH3 (g) + CO2 (g)  CO(NH2)2 (s) + H2O (l)

59. Example 4-15B
What mass of C6H11OH should you start with to produce 45.0 g C6H10 if the reaction has 86.2% yield and the C6H11OH is 92.3% pure: C6H11OH (l)  C6H H2O (l)

60. Exercise 26 Balance these equations by inspection
(NH4)2Cr2O7 (s)  Cr2O3 (s) + N2 (g) + H2O (g)
NO2 (g) + H2O (l)  HNO3 (aq) + NO (g)
H2S (g) + SO2 (g)  S (g) + H2O (g)
SO2Cl2 + HI  H2S + H2O + HCl + I2

61. Exercise 30 Write balanced equations for these reactions:
Sulfur dioxide gas with oxygen gas to produce sulfur trioxide gas
Solid calcium carbonate with water and dissolved carbon dioxide to produce aqueous calcium hydrogen carbonate
Ammonia gas and nitrogen monoxide gas to produce nitrogen gas and water vapor

62. Exercise 32 3 Fe (s) + 4 H2O (g)  Fe3O4 (s) + H2 (g)
How many moles of H2 can be produced from 42.7 g Fe and excess steam?
How many grams of H2O are consumed in the conversion of 63.5 g Fe to Fe3O4?
If 7.36 mol H2 are produced, how many grams of Fe3O4 must also be produced?

63. Exercise 36
Silver oxide decomposes above 300 °C to yield metallic silver and oxygen gas g impure silver oxide yields g O2. Assuming there is no other source of O2, what is the % Ag2O by mass in the original sample?

64. Exercise 42
How many grams of CO2 are produced in the complete combustion of 406 g of a bottled gas that consists of 72.7% C3H8 (propane) and 27.3% C4H10 (butane), by mass?

65. Exercise 45 What are the molarities of these solutes?
150.0 g sucrose (C12H22O11) in mL aqueous solution
98.3 mg of 97.9% pure urea, CO(NH2)2, in 5.00 mL aqueous solution
mL methanol (CH3OH, density = g/mL) in 15.0 L aqueous solution

66. Exercise 52
After 25.0 mL of aqueous HCl solution is diluted to mL, the concentration of the diluted solution is found to be M HCl. What was the concentration of the original HCl solution?

67. Exercise 56 Ca(OH)2 (s) + 2 HCl (aq)  CaCl2 (aq) + 2 H2O (l)
How many grams of Ca(OH)2 will react completely with 415 mL of M HCl?
How many kilograms of Ca(OH)2 will react with 324 L of an HCl solution that is 24.28% HCl by mass, density = 1.12 g/mL?

68. Exercise 63
g oxalic acid, H2C2O4, is exactly neutralized by mL of a NaOH solution. What is the concentration of the NaOH solution? H2C2O NaOH  Na2C2O H2O

69. Exercise 70
Chlorine can be generated by heating calcium hypochlorite and hydrochloric acid to form chlorine gas, calcium chloride, and water. If 50.0 g Ca(OCl)2 and 275 mL 6.00 M HCl react, how many grams of Cl2 gas form? Which reactant is left over, and how much (in grams)?

70. Exercise 72
2 C6H5NO2 + 4 C6H14O4  (C6H5N)2 + 4 C6H12O4 + 4 H2O
nitrobenzene triethylene azobenzene glycol
If 0.10 L nitrobenzene (d = 1.20 g/mL) react with 0.30 L triethylene glycol (d = 1.12 g/mL) to form 55 g azobenzene, find
Theoretical yield
Actual yield
Percent yield

71. Stoichiometry grams A grams B L of moles A moles B gas A gas B mL of
sol’n A
moles A
moles B
grams B
sol’n B
mole ratio
L of
gas A
gas B

72. Stoichiometry
grams A
mL A
moles A
moles B
grams B
mL B
mole ratio