1. Unit 5: Chemical Reactions

2. Law of Conservation of Mass
Atoms are neither created nor destroyed during a chemical reaction
Elements are present in the same amount before and after a reaction they simply move around, breaking old bonds and creating new ones

3. Chemical Reactions
Chemical reactions occur when two or more compounds interact, breaking bonds and creating new ones yielding products that are different from the substances being reacted

4. Chemical Reactions Many ways to know reaction has occurred
Temperature change
Colour change
Bubbles
Odour produced
Precipitate or new solid produced

5. Chemical Equations Used to summarize a chemical reaction
“+” means “reacts with”
“” means “produces”
Substances to the left of the arrow are reactants
Substances to the right of the arrow are products

6. Chemical Equations
Coefficients are the big numbers used to tell the ratio of mols of each reactant and product
Subscripts are small numbers used to tell the ratio of mols of each element

7. Chemical Equations

8. Chemical Equations
Subscripts are also used to show the state of reactants and products
(s) = solid
(l ) = liquid
(g) = gas
(aq) = aqueous (dissolved in water), used for solutions, acids, and bases

9. Balancing Chemical Equations
Remembering the Law of Conservation of Mass we have to have the same amount of each element is our products as we have in our reactants
We using “balancing” to make sure we follow this law

10. Balancing Chemical Equations
In reactions all subscripts make each compound what it is
We do not change small numbers in an equation
We change coefficients (big numbers) in order to balance equations
We have to balance equations so we have the lowest whole number ratio between all reactants and products

11. Balancing Chemical Equations
Balance the following equations:
H2(g) + O2(g)  H2O(l )
KClO3(aq)  KCl(s) + O2(g)
C2H6(g) + O2(g)  CO2(g) + H2O(g)

12. Writing Chemical Reactions
In a precipitation reaction, sodium hydroxide solution is mixed with iron (II) chloride solution. Sodium chloride solution and insoluble iron(II) hydroxide are produced. Write a balanced chemical equation including the state symbols.

13. Writing Chemical Reactions
Calcium hydroxide, a base reacts with phosphoric acid to yield solid calcium phosphate and liquid water. Determine the balanced equation.

14. Types of Reactions Combination Decomposition
Single displacement/replacement
Double displacement/replacement
Neutralization
Precipitate formation
Combustion

15. Combination Reactions
When two or more substances react to form one product
A + B  AB
For this course
Two elements combining
Formation of hydrates
Non-metal and metal oxides and water

16. Two Element Combination Reactions
Two elements combine to create a new compound
Predict the products of the following:
C(s) + O2(g) 
2Mg(s) + O2(g) 
N2(g) + 3H2(g) 

17. Hydrate Combination Reactions
Hydrates as stated before are chemical compounds that contain water or its parts
For this course we will mainly deal with hydrates of inorganic compounds (ionic compounds)
We can add water to compounds in a lattice work structure surrounding the central compound

18. Hydrate Combination Reactions
CoCl2 + 6H2O  CoCl2 •6H2O

19. Non-metal Oxide and Water Combination Reactions
Always forms an acid when oxide is soluble in water
Acids begin with H
Predict the products of the following:
SO2 + H2O 
Cl2O5 + H2O 

20. Metal Oxide and Water Combination Reactions
Always forms a base when oxide is soluble in water
Contain hydroxide ion (-OH)
Predict the products of the following:
Na2O + H2O 
MgO + H2O 

21. Decomposition Reactions
One substance undergoes a reaction to produce two or more other substances
AB  A + B
For this course:
Compound breaks down into its elements
Hydrates
Carbonates

22. Decomposition of Compound into its Elements
One compound (usually by heat) is forced to break down into stable states of the elements that make it up
Predict the products of the following:
HgO 
MgCl2 

23. Decomposition of Hydrates
We can heat hydrated substances to remove water molecules
The result is an “anhydrous” form of the compound
This can sometimes change the colour of the substance

24. Decomposition of Hydrates
CuSO4 •5H2O  CuSO4 + 5H2O

25. Decomposition of Carbonates
All carbonates break down to yield an oxide and CO2
Predict the products of the following:
CaCO3 
Na2CO3 

26. Single Displacement Reactions
AB + C  AC + B
We must use the activity series to determine whether a reaction will take place or not
The strongest/most reactive cation will create a bond
If the most reactive cation is already bonded then no reaction will take place

27. Single Displacement Reactions

28. Single Displacement Reactions
Those atoms closer to the top of the activity series (more reactive will replace lower elements in bonds.

29. Single Displacement Reactions
Determine whether the following reactions occur and write the products if a reaction is present:
AlCl3 + Ba 
ZnI2 + Pb 

30. Neutralization Reactions
Reactions involving acids and bases as the reactants
These reaction will yield water and a salt as their products
“salt” is any ionic compound whose cation comes from a base and anion comes from an acid

31. Neutralization Reactions
HCl(aq) + NaOH(aq)  H2O(l ) + NaCl(aq)

32. Double Displacement Reactions
AB + CD  AD + BC
Use solubility rules to determine if a precipitate is formed
No precipitate = no reaction
Precipitate is a solid that is form by two or more chemicals reacting together, sometimes changes the colour of the solution

33. Double Displacement Reactions

34. Double Displacement Reactions
Determine for the following if a reaction occurs, if so state the products and their states.
AgNO3(aq) + K3PO4(aq) 
CuF(aq) + AlCl3(aq) 
AlCl3(aq) + NaI(aq) 

35. Carbonates and Bicarbonates with Acids
React to form carbonic acid (H2CO3)
Carbonic acid is unstable and will continue to react
This gives H2O and CO2 as final products
HCl(aq) + NaHCO3(aq)  NaCl(aq) + H2CO3(aq)
H2CO3(aq)  H2O(l) + CO2(g)

36. Combustion Reactions
Combustion is burning (we always add O2(g) to whatever is being combusted)
In this course we assume there is always complete combustion
Combustion of hydrocarbons (compounds with only H and C) always produces CO2(g) and H2O(g)
All you need to do is balance (no guesswork with products)

37. Combustion Reactions
Predict the products of the following and balance the final equation
C4H10 + O2(g) 
C7H16 + O2(g) 

38. Net Ionic Equations For reactions in aqueous solution
Useful to show whether dissolved substances are present mostly as ions or molecules
Break down each reactant and product to the ions that make them up (if they are truly molecular and do not break up into ions write them as the whole compound)

39. Net Ionic Equations Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)
Becomes:
Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2(s) + 2K+(aq) + 2NO3-(aq)
The above is the complete ionic equation
All soluble strong electrolytes are shown as ions
Remember if they aren’t soluble they are left
PbI2(s)

40. Net Ionic Equations
Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2(s) + 2K+(aq) + 2NO3-(aq)
K+(aq) and NO3-(aq) appear on both sides of the equation in the same form
These are called spectator ions
Don’t participate in the reaction
When spectator ions are omitted we have the net ionic equation
Pb2+(aq) + 2I-(aq)  PbI2(s)

41. Net Ionic Equations Write Net Ionic Equations for the following
AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)
Mg(NO3)2(aq) + Na2CO3(aq)  MgCO3(s) + NaNO3(aq)

42. Oxidation Reduction Reactions
Redox reactions
Reactions where electrons are transferred from one reactant to another

43. Redox Reactions LEO the lion says GER Loss of Electrons is Oxidation
Gain of Electrons is Reduction
When one substance is oxidized another substance must be reduced
Electrons have to go somewhere

44. Redox Reactions
Each atom in a neutral substance of ion is assigned an oxidation number (oxidation state)
Oxidation states allow us to keep track of electrons being gained or lost
Oxidation numbers range from -4 to +7

45. Oxidation Numbers
Oxidation numbers of atoms in covalently bonded compound depend on electronegativity
More electronegative atoms carrying a more negative oxidation state
Written with +/- before the number (S-2)
Charge is written +/- after the number (S2-)

46. Oxidation Numbers
1. Atoms in their natural state have oxidation number of 0
Each H in H2 has oxidation state of 0
Each S in S8 has oxidation state of 0
Each P in P4 has oxidation state of 0
Solid metals (ie. Fe(s))have oxidation state of 0

47. Oxidation Numbers
2. Monoatomic ions’ oxidation number = ionic charge K+ = oxidation state +1 S2- = oxidation state -2

48. Oxidation Numbers 3. Non-metals usually have negative oxidation state
Some exceptions
O usually -2
Exception = peroxides (contain O22- ion, given -1)
H usually +1 with non-metals, -1 with metals
F is -1 in all compounds
Other halogens = -1 in most binary compounds
When combined with O have positive oxidation states

49. Oxidation Numbers Sum of oxidation numbers in neutral compound = 0
Sum of oxidation numbers in polyatomic ion =charge of the ion
H3O+ = 3(+1) + 1(-2) = +1 = charge of ion

50. Oxidation Numbers
What is the oxidation number of sulfur in the following:
H2S S8
SCl2
Na2SO3
SO42-

51. Redox Reaction
In order for a reaction to be considered redox there needs to be a transfer of electrons
This means that there must be a change in oxidation state
Remember, when an electron is lost by one substance another substance must gain that electron
We cannot have oxidation without reduction

52. Finding Redox Reactions
Determine which of the following are redox:
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
Na2CO3(s) + 2HCl(aq)  2NaCl(aq) + H2O(l) + CO2(g)

53. Redox Agents Oxidizing agent undergoes reduction
Causes another species to be oxidized
Reducing agent undergoes oxidation
Causes another species to be reduced

54. Redox Agents
Determine the oxidizing agent and reducing agent in each example:
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

55. Stoichiometry
We can use balanced chemical equations to convert from one substance to another
This can help us determine how much of our product we can expect to get
Remember, equations must be balanced!!!!

56. Stoichiometry Coefficients in an equation tell us the mol ratio
Compares mols of one substance to another
To convert between two different substances we must be in mols

57. Stoichiometry
Given the following equation how many mols of ZnBr2 will be produced from mols of solid Zn?
Zn(s) + 2HBr(aq)  ZnBr2(aq) + H2(g)

58. Stoichiometry
Given the following equation how many mols of solid Pb will be produced from 33.3gPb(NO3)2?
Mn(s) + Pb(NO3)2(aq)  Mn(NO3)2(aq) + Pb(s)

59. Stoichiometry
Given the following equation how many grams of H2 can be made with 6.5gHCl?
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

60. Limiting Reagents (Reactants)
Imagine you are making sandwiches using bread and cheese
For every 2 pieces of bread you need one piece of cheese
Bd = bread
Ch = cheese
Bd2Ch = sandwich

61. Limiting Reagents (Reactants)
The following equation represents your sandwich making:
2Bd + Ch  Bd2Ch
If you have 10 slices of bread and 7 slices of cheese you would only be able to make 5 sandwiches
Once you run out of bread you can’t make any more

62. Limiting Reagents (Reactants)
In the lab once we run out of one of our reactants we can’t make any more products
The first reactant to run out is known as the limiting reagent or limiting reactant
In calculations we use the limiting reagent for calculating the amount of product that will be made
Remember to balance your equation!

63. Limiting Reagents (Reactants)
Given the following reaction how many mols NH3 can be formed from 3.0mol N2 and 6.0 mol H2?
N2(g) + 3H2(g)  2NH3(g)

64. Limiting Reagents (Reactants)
How many grams of H2O can be formed according to the following equation if 150g H2 are reacted with 1500g O2?
2H2(g) + O2(g)  2H2O(g)

65. Percent Yield Comparison between theoretical yield and actual yield
Theoretical yield = calculated amount of product
From amount of reactants started with (remember use limiting reagent)
Actual yield = from doing experiment
Must be given in problem or found in lab

66. Percent Yield
%𝑦𝑖𝑒𝑙𝑑= 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 x 100%

67. Percent Yield
Student preformed the following experiment in lab. Using 3.00gTi and 6.00gCl2 students were able to produce 7.7gTiCl4. Determine the percent yield for this experiment.
Ti(s) + Cl2(g)  TiCl4(l)