1. CLASSIFYING CHEMICAL REACTIONS
Review of Writing and Balancing Pg

2. Review:
When counting elements, don’t forget to look at both the subscript and the coefficient.
For example:
P2O5 = has 2 phosphorus atoms and 5 oxygen atoms
2P2O5 = has 4 phosphorus atoms and 10 oxygen atoms
Because there are 2 molecules (indicated by the coefficient) and 2 atoms in each molecule (indicated by the subscript) –
So you multiply!!

3. Remember: 2O2(g) + 4H2(g)  4H2O(l)
Never change a subscript to balance an equation!
O2(g) + H2(g)  H2O(l) Is unbalanced – but you can’t change it the following way!
O2(g) + H2(g)  H2O2(l) X
Make sure the coefficients are the lowest whole-number ratio :
2O2(g) + 4H2(g)  4H2O(l)
This is a balanced formula but these are not the lowest numbers you could use:
O2(g) + 2H2(g)  2H2O(l)

4. Balancing Chemical Equations
1) Write the chemical formulas for the reactants and products including the states
Cu(s) + AgNO3(aq)  Ag(s) + Cu(NO3)2(aq)
Balance the element (atom or ion) present in the greatest number by multiplying by the lowest coefficient possible
(NO3)2(aq) = 2 present (lowest coefficient possible to balance = 2)
Cu(s) AgNO3(aq)  Ag(s) + Cu(NO3)2(aq)
3) Repeat step 2 for the rest of the elements
Now we have 2 Ag, so balance the other side
Cu(s) + 2AgNO3(aq)  2Ag(s) + Cu(NO3)2(aq)
4) Count elements on each side of the final equation to ensure they balance:
1 Cu(s) = 1 Cu(s) ; 2Ag = 2Ag(s) ; 2 NO3 = (NO3)2(aq)

5. What do you remember?
You learned about five reaction types, can you match them up
Composition (Formation)
Decomposition
Combustion
Single Replacement
Double Replacement
CH4(g) + O2(g)  CO2(g) + H2O(g)
Mg(s) + O2(g)  MgO(s)
Cu(s) + AgNO3(aq)  Ag(s) + Cu(NO3)2(g)
CaCl2(aq) + Na2CO3(aq)  CaCO3(s) + NaCl(aq)
H2O(l)  O2(g) + H2(g)

6. Classifying Chemical Reactions: Composition (Formation)
2Mg(s) + O2(g)  2MgO(s)
element + element  compound
Predict and balance the following:
Li(s) + Cl2(g) 
Na(s) + F2(g) 
Ba(s) + N2(g) 

7. Classifying Chemical Reactions: Decomposition
2H2O(l)  O2(g) + 2H2(g)
compound  element + element
Predict and balance the following:
NaCl(s) 
Sr3P2(s) 
Cs2O(s) 

8. Classifying Chemical Reactions: Combustion
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
fuel + oxygen  carbon dioxide + water
Predict and balance the following:
C3H8(g) + O2(g) 
C2H6(g) + O2(g) 
C4H10(g) + O2(g) 

9. Classifying Chemical Reactions: Single Replacement
2AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + 2Ag(s)
compound + element  compound + element
2NaI(aq) + Cl2 (g)  I2 (s) + 2NaCl(aq)
Predict and balance the following:
NaBr(aq) + O2(g) 
CuCl2(aq) + Al(s) 
Li2CO3(aq) + K(s) 

10. Classifying Chemical Reactions: Double Replacement
2AgNO3(aq) + CuCl2(aq)  Cu(NO3)2(aq) + 2AgCl(aq)
compound + compound  compound + compound
CaCI2(aq) + Na2CO3 (aq)  2NaCl (aq) + CaCO3(s)
Predict and balance the following:
NaBr(aq) + MgO(aq) 
CuCl2(aq) + AlF3(aq) 
Li2CO3(aq) + K2O(aq) 

11. Using the solubility table:

12. Chemical Reaction Equations
Section 7.1 pg

13. Reaction Assumptions
Reactions are spontaneous – reactions will occur when all the reactants are mixed together
Reactions are fast – the reaction must occur within a reasonable time (see pg. 280)
Reactions are quantitative – one that is more than 99% complete; in other words, at least one reactant is completely used up
Reactions are stoichiometric – means that there is a simple whole-number ratio of chemical amounts of reactants and products (the coefficients for a balanced equation do not change)

14. Net Ionic Equations
A chemical reaction equation that includes only reacting entities (molecules, atoms and/or ions) and omits any that do not change
Writing Net Ionic Equations:
1) Write a complete balanced chemical equation
2) Dissociate all high-solubility ionic compounds, and ionize all
strong acids to show the complete ionic equation
3) Cancel identical entities that appear on both the reactant and
product sides
4) Write the net ionic equation, reducing coefficients if
neccessary

15. (Complete ionic equation)
Practice
When cancelling spectator ions, they must be identical in every way: chemical amount, form (atom, ion, molecule) and state of matter
Write the net ionic equation for the reaction of aqueous barium chloride and aqueous sodium sulfate. (Refer to the solubility table)
1) BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)
2) Ba2+(aq) + 2Cl-(aq) +2Na+(aq) + SO42-(aq)  BaSO4(s) + 2Na+(aq) + 2Cl-(aq)
(Complete ionic equation)
3) Ba2+(aq) + 2Cl-(aq) +2Na+(aq) + SO42-(aq)  BaSO4(s) + 2Na+(aq) + 2Cl-(aq)
4) Ba2+(aq)) + SO42-(aq)  BaSO4(s) (Net ionic equation)
Ions that are present but do not take part in (change during) a reaction are called spectator ions (like spectators at a sports game: they are present but do not take part in the game)

16. Practice
Write the net ionic equation for the reaction of zinc metal and aqueous copper (II) sulfate (Refer to the solubility table)
1) Zn(s) + CuSO4(aq)  Cu(s) + ZnSO4(aq)
2) Zn(s) + Cu2+(aq) + SO42-(aq)  Cu(s) + Zn2+(aq) + SO42-(aq)
(Complete ionic equation)
3) Zn(s) + Cu2+(aq) + SO42-(aq)  Cu(s) + Zn2+(aq) + SO42-(aq)
4) Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq) (Net ionic equation)

17. Practice
For entities involving strong acids, H+(aq) is used as a matter of convenience over H3O+(aq)
Write the net ionic equation for the reaction of hydrochloric acid and barium hydroxide
1) 2HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2HOH(l)
aka: 2H2O(l)
2) 2H+(aq) + 2Cl-(aq) + Ba2+(aq) + 2OH-(aq)  Ba2+(aq) + 2Cl-(aq) + 2H2O(l)
(Complete ionic equation)
3) 2H+(aq) + 2Cl-(aq) + Ba2+(aq) + 2OH-(aq)  Ba2+(aq) + 2Cl-(aq) + 2H2O(l)
4) H+(aq)) + OH-(aq)  H2O(l) (Net ionic equation) – coefficients reduced to 1

18. Limiting and Excess Reagents
Cu(s) + AgNO3(aq) 
What is in the container when the reaction is finished?
 Cu(NO3)2(aq) + Ag(s) (Cu2+(aq) = blue)
But how do you know the reaction is done?? There is still copper left!
(a) Copper wire and a beaker with aqueous silver nitrate solution
(b) A few moments after the wire is immersed
(c) The beaker contents after 24 h

19. Limiting and Excess Reagents
When no further changes appear to be occurring, we assume that all of the AgNO3(aq) that was initially present has now been completely reacted.
A limiting reagent is the reactant whose entities are completely consumed in a reaction, meaning the reaction stops.
In order to make sure this happens, more of the other reactant must be present than is required
An excess reagent is the reactant whose entities are present in surplus amounts, so that some remain after the reaction ends..
In our reaction: much more copper was used than needed (evidenced by the unreacted copper) so we assume the reaction ended when no more silver ions were left, so silver nitrate was the limiting reagent.

20. Homework: WB pg 59 Q 10 WB pg 71 Q 42 WB pg 91 Q 1-3
* Classifying & Balancing Equations WS

21. Stoichiometry
Section 7.2 pg

22. Stoichiometry
A method of problem-solving using known quantities of one entity in a chemical reaction to find out unknown quantities of another entity in the same reaction
Always requires a balanced chemical equation
Series of steps for measurement calculations
Always involves a mole ratio to “switch substances”
Writing chemical reactions, net ionic equations and dissociation equations is essential
Converting grams to moles (and vice versa) is also essential

23. Gravimetric Stoichiometry
Gravimetric Stoichiometry – method used to calculate the masses of reactants or products in a chemical reaction
Gravimetric = mass measurement
Using gravimetric stoichiometry, we can apply our knowledge of balanced chemical reactions and mass to mol conversions to:
Predict the mass of product we will get from a reaction
Estimate the amount of reactant we need for a reaction to produce a certain mass of product

24. Gravimetric Stoichiometry Steps:
Step 1: Write a balanced chemical reaction equation
List the measured mass (given), the unknown quantity (required) and conversion factors (M = molar mass)
Step 2: Convert the mass of the measured substance to moles
Step 3: Calculate the moles of the unknown substance using the
mole ratio required
given
Step 4: Convert from moles of the required substance to its mass.

25. Stoichiometry Calculations
(Measured quantity) solids/liquids m  n (Required quantity)
mole ratio

26. The balanced chemical equation is:
How many grams of oxygen are required to completely burn 10.0 g of propane (C3H8(g))?
The balanced chemical equation is:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
The number of moles of propane reacting is: (MASS to MOLES)
10.0 g x 1 mol = mol
44.11 g
The number of moles of oxygen produced is: (MOLE RATIO)
0.227 mol x 5 mol = mol
1 mol
The mass of oxygen produced is: (MOLES TO MASS)
mol x g = g

27. Practice #2
20.0 g of butane (C4H10(g)) are completely burned in a lighter. How many grams of CO2(g) are produced?
Start with a balanced chemical equation:
2 C4H10(g) O2(g)  8 CO2(g) H2O (g)
m = 20.0g m = ?
M = g/mol M = g/mol

28. Practice #2 (Team Unit Analysis)
2 C4H10(g) + 13 O2(g)  8 CO2(g) H2O (g)
m = 20.0g m = ?
M = g/mol M = g/mol
2) Mass to moles, mole ratio, moles to mass
20.0 g x mol x mol CO x g = g
58.14 g mol C4H mol

29. Practice #3 (Unit Analysis)
What mass of iron (III) oxide is required to produce g of iron?
Fe2O3(s) CO(g)  2 Fe(s) + 3 CO2(g)
m = ? m = 100.0g
M = g/mol M = g/mol
m Fe2O3(s): g x mol x 1 mol x g = g Fe2O3
55.85 g mol mol

30. Practice
Remember to keep the unrounded values in your calculator for further calculation until the final answer is reported.
WB pg ALL
Pg. 290 #8-14
(Answers pg. 785)

31. Percent Yield for Reactions
We can use stoichiometry to test experimental designs, technological skills, purity of chemicals, etc. We evaluate these by calculating a percent yield.
This is the ratio of the actual (experimental) quantity of product obtained and the theoretical (predicted) quantity of product obtained from a stoichiometry calculation
Percent yield = actual yield x
predicted yield
Some forms of experimental uncertainties:
All measurements (limitations of equipment)
Purity of chemical used ( % purity)
Washing a precipitate (some mass is lost through filter paper)
Estimation of reaction completion (qualitative judgements i.e. color)

32. Percent Yield Example #1
Example: In a chemical analysis, 3.00 g of silver nitrate in solution was reacted with excess sodium chromate to produced 2.81 g of filtered, dried precipitate.
Predicted value: 2AgNO3(aq) + Na2CrO7(aq) Ag2CrO7(s) +2NaNO3(aq) m = 3.00 g m = M = g/mol M = g/mol
3.00g x mol x mol x g = g
169.88g mol mol
Percent yield = actual yield x % = 2.81g x 100% = %
predicted yield g

33. Testing the Stoichiometric Method
Stoichiometry is used to predict the mass of precipitate actually produced in a reaction.
Filtration is used to separate the mass of precipitate actually produced in a reaction

34. Testing the Stoichiometric Method
What mass of lead is produced by the reaction of 2.13 g of zinc with an excess of lead(II) nitrate solution?
Design: A known mass of zinc is place in a beaker with an excess of lead(II) nitrate solution. The lead is produced in the reaction is separated by filtration and dried. The mass of the lead is determined
Prediction: Zn(s) Pb(NO3)2(aq)  Zn(NO3)2(aq) + Pb(s)
m = 2.13 g m = ?
M = g/mol M = g/mol
2.13 g x 1 mol x x g = g
65.41 g mol

35. Testing the Stoichiometric Method
Prediction: g of lead will be produced (Stoichiometric calculation)
Evidence: In the beaker, crystals of a shiny black solid were produced, and all the zinc disappeared.
Mass of filter paper = 0.92 g Mass of dried filter paper plus lead = 7.60g
Analysis:
mass of lead product = 7.60g – 0.92g = 6.68g
According to the evidence collected, 6.68 g of lead precipitated.
Evaluation: % difference = (experimental – predicted) x 100 %
predicted
6.68 – x 100% = 1%
6.75
Given such a small difference, which can readily be accounted for by normal sources of error, the prediction is clearly verified, and so the stoichiometric method for predicting the mass of lead formed in this experiment is judged to be acceptable.

36. Testing the Stoichiometric Method
Prediction: g of lead will be produced (Stoichiometric calculation)
Analysis:
mass of lead product = 7.60g – 0.92g = 6.68g
According to the evidence collected, 6.68 g of lead precipitated.
Evaluation:
Could you also calculate the % yield ?
Percent yield = actual yield x % = g x 100% = 99.0%
predicted yield g

37. Homework
Pg. 293 #1-2, 6-10

38. Gas Stoichiometry
Section 7.3 pg

39. Molar Volume
STP = 22.4L/mol
SATP = 24.8 L/mol
Molar volume is the same for all gases at the same temperature and pressure (remember, all gases have the same physical properties)
At STP, molar volume = 22.4 L/mol ( kPa and 0oC)
At SATP, molar volume = 24.8 L/mol (100 kPa and 25oC)
This can be used as a conversion factor just like molar mass!
At STP, one mole of gas has a volume of 22.4 L, which is approximately the volume of 11 “empty” 2 L pop bottles.

40. Comparing Kelvin and Celsius Scales
To convert degrees Celsius to Kelvin , you add 273.
K = oC + 273
To convert Kelvin to degrees Celsius, you subtract 273.
oC = K - 273
Examples:
What is 254 K in oC ?
-19oC
What is -34oC in K ?
239K

41. Gas Stoichiometry
Many chemical reactions involve gases as a reactant or a product
Gas Stoichiometry – the procedure for calculating the volume of gases as products or reactants
Gases also have a molar volume (L/mol) rather than concentration.
This is the conversion factor used to convert (litres of gas) to (moles of gas)
The Ideal Gas Law (PV = nRT) may also be required to:
A) find the number of moles of reactant
B) Find the V, P, or T of the product

42. Example #1
If 300g of propane burns in a gas barbecue, what volume of oxygen measured at SATP is required for the reaction?
Remember: 24.8L/mol for SATP
C3H8(g) O2(g)  3CO2(g) H2O(g)
m = 300g V = ?
44.11g/mol 24.8L/mol
300 g x 1 mol x 5 mol x L = 843 L O2(g)
44.11 g mol mol
**Remember – molar volume is the conversion factor for gases just like molar mass is the conversion factor in gravimetric stoichiometry

43. Example #2
Hydrogen gas is produced when sodium metal is added to water. What mass of sodium is necessary to produce 20.0L of hydrogen at STP?
Remember: 22.4L/mol for STP
2Na(s) H2O (l)  2NaOH(aq) + H2(g)
m = ? V = 20.0L
22.99g/mol L/mol
20.0L x 1 mol x 2mol x g = 41.1 g Na(s)
22.4 L mol mol
**Remember – molar volume is the conversion factor for gases just like molar mass is the conversion factor in gravimetric stoichiometry

44. Example #3 2N2(g) + 3H2(g)  2NH3(g) m = 7500g m = ?
If the conditions are not STP or SATP, the molar volume cannot be used! You must use the ideal gas law to find the gas values using moles determined from stoichiometry
What volume of ammonia at 450kPa and 80oC can be obtained from the complete reaction of 7.5kg of hydrogen with nitrogen?
2N2(g) H2(g)  NH3(g)
m = 7500g m = ?
M = 2.02 g/mol P = 450kPA
T = K
7500 g x 1 mol x = mol NH3(g)
2.02 g
PV = nRT  V = nRT = ( mol)(8.314kpa•L/mol•K)(353.15K)
P (450kPa)
= L  x 104 L of NH3(g)

45. Gas Stoichiometry Summary
Write a balanced chemical equation and list the measurements, unknown quantity symbol, and conversion factors for the measured and required substances.
Convert the measured quantity to a chemical amount using the appropriate conversion factor
Calculate the chemical amount of the required substance using the mole ratio from the balanced chemical equation.
Convert the calculated chemical amount to the final quantity requested using the appropriate conversion factor.

46. Stoichiometry Calculations
(Measured quantity) solids/liquids m  n gases V, T, P gases V,T,P solids/liquids m  n (Required quantity)
mole ratio

47. Homework
Gas Stoichiometry Worksheet
Pg. 299 #1-6

48. Solution Stoichiometry
Section 7.4 pg

49. Solution Stoichiometry
The majority of work in research and industry involves solutions. Recall that solutions are easy to handle and are usually easier to control in reactions.
The major difference compared to gravimetric and gas stoichiometry is that we use molar concentration (mol/L) as a conversion factor rather than molar mass or molar volume
Solution stoichiometry – the procedure for calculating the molar concentration or volume of solution products or reactants

50. Example #1
Ammonia and phosphoric acid solutions are used to produce ammonium hydrogen phosphate fertilizer. What volume of 14.8mol/L NH3(aq) is needed to react with 1.00kL of 12.9mol/L of H3PO4(aq)?
2NH3(aq) H3PO4(aq)  (NH4)2HPO4(aq)
V = ? V = 1.00kL
14.8mol/L mol/L
1.00kL x mol x 2mol x 1 L = 1.74 kL
1 L mol mol = 1.74 x 103 L

51. Example #2 H2SO4(aq) + 2KOH(aq)  2H2O(l) + K2SO4(aq)
In an experiment, a mL sample of sulfuric acid solution reacts completely with 15.9 mL of mol/L potassium hydroxide. Calculate the amount concentration of the sulfuric acid.
H2SO4(aq) KOH(aq)  H2O(l) + K2SO4(aq)
V = 10.00mL V = 15.9 mL
c = ? mol/L
15.9mL x mol x 1mol x = mol/L
1 L mol mL

52. Gravimetric, Gas and Solution Stoichiometry Summary
Write a balanced chemical equation and list the quantities and conversion factors for the given substance and the one to be calculated.
Convert the given measurement to its chemical amount using the appropriate conversion factor
Calculate the amount of the other substance using the mole ratio from the balanced chemical equation.
Convert the calculated chemical amount to the final quantity requested using the appropriate conversion factor.
Remember to use the Ideal Gas Law for all gases not at STP or SATP

53. Stoichiometry Calculations
(Measured quantity) solids/liquids m  n gases V, T, P solutions c, V solutions c, V gases V,T,P solids/liquids m  n (Required quantity)
mole ratio

54. Guided Practice
WB pg ALL Questions