1. By Professor Walker Waugh
Balancing Reactions By
Professor Walker Waugh

2. LAW OF CONSERVATION OF MASS
Balancing Reactions
Remember the
LAW OF CONSERVATION OF MASS
Matter CANNOT be created nor destroyed!
So USE COMMON SENSE:
However many atoms you have on the reactant side you must also have on the product side

3. Method for Balancing Reactions
Treat all polyatomic ions as ONE UNIT, if it does not change from reactant to product
Balance all NON-hydrogen and oxygen atoms
Balance all hydrogen atoms
Balance all oxygen atoms
Check yourself before you WRECK yourself!

4. Simple things to watch for when Balancing Reactions
Remember you can change the coefficients but NOT the subscripts
Combustion reactions means that when a substance is burned it reacts with O2 (g)
Oxygen MUST be present for something to burn, it supports combustion!
Odd number of oxygen atoms on the product side
Oxygen atoms in the reactant being burned

5. Balancing Pitfall #1 Odd number of oxygen atoms on the product side
Butane gas (C4H10) burns
C4H10 (g) + O2 (g)  CO2 (g) + H2O (g)
NOT H2O (l)
1. Balance NON-hydrogen & oxygen atoms
C4H10 (g) + O2 (g)  4 CO2 (g) + H2O (g)
four C four C
2. Balance hydrogen atoms
C4H10 (g) + O2 (g)  4 CO2 (g) H2O (g)
ten H H
(because the coefficient 5 times the subscript 2 = 10)
3. Balance oxygen atoms
two O eight O + five O = thirteen O
4. Make a fraction because 13/2 x 2 = 13
C4H10 (g) + 13/2 O2 (g)  4 CO2 (g) H2O (g)

6. Balancing Pitfall #1 Final Answer:
Multiply the ENTIRE equation by 2 to remove fraction (we don’t have partial compounds)
2 (C4H10 (g) + 13/2 O2 (g)  4 CO2 (g) H2O (g))
Final Answer:
2 C4H10 (g) O2 (g)  8 CO2 (g) H2O (g)
Check It
8 C 8
20 H 20
26 O = 26

7. Balancing Pitfall #2 Oxygen atoms in the burning reactant
CH3OH (g) + O2 (g)  CO2 (g) + H2O (g)
Same rules as before but subtract an oxygen atom from the reactant side before you try to balance O2
CH3OH (g) + 3/2 O2 (g)  CO2 (g) + 2H2O (g)
Again: Multiply by 2 to remove the fraction

8. 2 CH3OH (g) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g)
Balancing Pitfall #2
2( CH3OH (g) + 3/2 O2 (g)  CO2 (g) + 2H2O (g))
Final Answer:
2 CH3OH (g) O2 (g)  2 CO2 (g) H2O (g)
Check It
2 C 2
8 H 8
2 + 6 = 8 O = 8

10. By Professor Walker Waugh
Double Displacement Reactions Precipitation, Neutralization & Gaseous Reactions By
Professor Walker Waugh

11. Predicting Products: Double Displacement Reactions
There are three types of Double Displacement Reactions
Precipitation Reactions
Acid/Base Neutralizations
Gaseous Reactions
All occur via the same reaction mechanism:
Swinger’s Mechanism
A+B− + C+D−  A+D− + C+B−
Remember to write the Cations (+) first then the Anions (-) in each compound!

12. Predicting Products: Precipitation Reactions
Method:
Recognize the reaction type and write the correct formulas of the reactants:
Salt + Salt
Salt + Acid
Salt + Base
Swing the cations.
Check the formulas of the products, make sure that the nomenclature is correct.
Balance the reaction.
Check the Solubility Table to see if there is a solid formed.
Salt: A classification of chemical compounds
A salt is any aqueous ionic compound that is NOT an acid (e.g. HCl(aq) ), base (e.g. NaOH) or oxide (e.g. Li2O = Lithium oxide)

13. Predicting Products: Precipitation Reactions
Example
Aqueous barium nitrate is placed in a test tube containing aqueous lithium phosphate
(I don’t give products for Precipitation Reactions)
Ba(NO3)2 (aq) + Li3PO4 (aq)  BaPO4 ___ + LiNO3 ___
BaPO4 is NOT a correct formula Ba3(PO4)2 is the correct formula
LiNO3 is the correct formula
Re-examine the formula and balance it, including correct product formulas:
3 Ba(NO3)2 (aq) + 2 Li3PO4 (aq)  Ba3(PO4)2 ___ + 6 LiNO3 ___

14. Predicting Products: Precipitation Reactions
Finally, check the Solubility Table, located on last page of Reaction Table 100, to assign the states to the products.
If the table says that the substance is INSOLUBLE then it does NOT dissolve in water = (s)
If the Table says that the substance in SOLUBLE then it DOES dissolve in water = (aq)
Make sure to read the EXCEPTIONS!
For Ba3(PO4)2 see Rule #5
For LiNO3 see Rule #1
Final Answer
3 Ba(NO3)2 (aq) + 2 Li3PO4 (aq)  Ba3(PO4)2 (s) + 6 LiNO3 (aq)

15. Solubility Table: know if taking General Chemistry next semester
SOLUBILITY RULES FOR IONIC COMPOUNDS
Compounds containing Group IA metals, ammonium, acetates and nitrates are all soluble.
Most halides (Group 7A - chlorides etc.) are soluble. Exceptions include Ag+1, Pb+2, and Hg2+2 halides.
Most sulfates are soluble. Exceptions include Ba+2, Sr+2, Ag+1, Pb+2, and Ca+2 sulfates.
Most hydroxides insoluble. Exceptions include hydroxides of Group 1A metals, ammonium, Ca+2, Sr+2, and Ba+2.
Most phosphates, carbonates, chromates, and sulfides are insoluble. Exceptions include those compounds containing Group 1A metals and ammonium.
In addition, all acids are soluble!

16. Predicting Products: Acid/Base Neutralization Reactions
Method:
Recognize the reaction type and write the correct formulas of the reactants:
Acid + Base (ONLY as Reactants)
Swing the cations.
When H+ + OH− combines then make H-OH (which is H2O (l) ).
Check the formulas of the products, make sure that the nomenclature is correct.
Refer to the Solubility Table and assign states of products.
Balance the reaction.
Check the Solubility Table to see if there is a solid formed.

17. Predicting Products: Acid/Base Neutralization Reactions
Example
Hydrobromic acid is placed in a test tube containing aqueous calcium hydroxide
HBr (aq) + Ca(OH)2 (aq)  CaBr___ H-OH___
HBr (aq) + Ca(OH)2 (aq)  CaBr2___ H-OH___
HBr (aq) + Ca(OH)2 (aq)  CaBr2 (aq) + H-OH (l)
2 HBr (aq) + Ca(OH)2 (aq)  CaBr2 (aq) H-OH (l)
It is easier to balance the reaction when water is written as as H-OH in acid/base neutralization reactions.
Final Answer
2 HBr (aq) + Ca(OH)2 (aq)  CaBr2 (aq) H2O (l)

18. Predicting Products: Gaseous Reactions
There are four types of Gaseous Displacement Reactions that you must know!
Carbonate (CO32-, or HCO3-) + Acid Reactions
Sulfites (SO32-, or HSO3-) + Acid Reactions
Sulfide (S2-) + Acid Reactions
Ammonium ion (NH4+) + Hydroxide (OH-) Reactions
All occur via the same reaction mechanism:
Swinger’s Mechanism
A+B− + C+D−  A+D− + C+B−

19. Predicting Products: Gaseous Carbonate Reactions
Method:
Recognize the reaction type and write the correct formulas of the reactants:
CO Acid
HCO3− + Acid
Swing the cations.
When H2CO3(aq), carbonic acid, is formed then it automatically dissociates to H2O (l) and CO2 (g)
Check the formulas of the products, make sure that the nomenclature is correct.
Balance the reaction.
Check the Solubility Table to see if there is a solid formed.

20. Predicting Products: Gaseous Carbonate Reactions
Example
Hydrochloric acid is placed in a beaker containing aqueous sodium carbonate
HCl (aq) + Na2CO3 (aq)  NaCl (aq) + HCO3 (aq)
HCO3 (aq) is NOT a good formula. H2CO3 (aq) = carbonic acid
HCl (aq) + Na2CO3 (aq)  NaCl (aq) + H2CO3 (aq)
Balance the formula since all formulas are correct:
2 HCl (aq) + Na2CO3 (aq)  2 NaCl (aq) + H2CO3 (aq)
Remember that H2CO3 (aq)  H2O (l) + CO2 (g)
Final Answer
2 HCl (aq) + Na2CO3 (aq)  2 NaCl (aq) + H2O (l) + CO2 (g)

21. Predicting Products: Gaseous Sulfite Reactions
Method:
Recognize the reaction type and write the correct formulas of the reactants:
SO Acid
HSO3− + Acid
Swing the cations.
When H2SO3(aq), sulfurous acid, is formed then it automatically dissociates to H2O (l) and SO2 (g)
Check the formulas of the products, make sure that the nomenclature is correct.
Balance the reaction.
Check the Solubility Table to see if there is a solid formed.

22. Predicting Products: Gaseous Sulfite Reactions
Example
Nitric acid is placed in a beaker containing aqueous sodium hydrogen sulfite
HNO3 (aq) + NaHSO3 (aq)  NaNO3 (aq) + H-HSO3 (aq)
H-HSO3 (aq) is really H2SO3 (aq) = sulfurous acid.
Remember that H2SO3 (aq)  H2O (l) + SO2 (g)
Balance the formula since all formulas are correct and check the Solubility Table for state of the products:
Final Answer
HNO3 (aq) + NaHSO3 (aq)  NaNO3 (aq) + H2O (l) + SO2 (g)

23. Predicting Products: Gaseous Sulfide Reactions
Method:
Recognize the reaction type and write the correct formulas of the reactants:
S Acid
Swing the cations.
When H2S results it is also a gaseous product.
Check the formulas of the products, make sure that the nomenclature is correct.
Balance the reaction.
Check the Solubility Table to see if there is a solid formed.

24. Predicting Products: Gaseous Sulfide Reactions
Example
Hydrochloric acid is placed in a beaker containing aqueous lithium sulfide
HCl (aq) + Li2S (aq)  LiCl___ + HS___
HS is NOT a good formula. H2S is hydrogen sulfide.
Remember that H2S when formed is GASEOUS and smells like ROTTEN EGGS!
Balance the formula since all formulas are correct and check the Solubility Table for state of the products:
2 HCl (aq) + Li2S (aq)  2 LiCl (aq) + H2S (g)
Final Answer
Same as Above

25. Predicting Products: Gaseous Ammonium Reactions
Method:
Recognize the reaction type and write the correct formulas of the reactants:
NH OH−
Swing the cations.
When NH4OH (aq), ammonium hydroxide, is formed then it automatically dissociates to H2O (l) and NH3 (g)
Check the formulas of the products, make sure that the nomenclature is correct.
Balance the reaction.
Check the Solubility Table to see if there is a solid formed.

26. Predicting Products: Gaseous Ammonium Reactions
Example
Aqueous ammonium chloride is placed in a beaker containing aqueous barium hydroxide
NH4Cl (aq) + Ba(OH)2 (aq)  BaCl (aq) + NH4OH (aq)
BaCl (aq) is NOT the correct formula for barium chloride.
Barium chloride (aq) = BaCl2.
Thus: 2 NH4Cl (aq) + Ba(OH)2 (aq)  BaCl2 (aq) NH4OH (aq)
Remember that NH4OH = NH3 (g) + H2O (l)
Final Answer
2 NH4Cl (aq) + Ba(OH)2 (aq)  BaCl2 (aq) NH3 (g) H2O (l)

28. Molecular, Ionic and Net Ionic EquationsBy
Professor Walker Waugh

29. Molecular, Ionic and Net Ionic Equations
Molecular Equations: We will write all compounds as molecules even if ionic
Complete Ionic Equations (CIE): Write all ionic compounds and strong acids as ions. Do not break up precipitants or weak acids on product side.
Know the 7 strong acids (any thing else is considered a weak acid)
HCl (aq), HBr (aq), HI (aq), HNO3 (aq), H2SO4 (aq), HClO4 (aq), HClO3 (aq) (chloric acid is NOT noted in your textbook)
Net Ionic Equations (NIE): Complete Ionic Equations – Spectator Ions
Spectator Ions = any ion that does NOT change in any way from reactant to product side of the equation

30. Molecular, Ionic and Net Ionic Equations
Molecular Equations =
A Balanced Chemical Equation
Example #1
Sulfuric acid is placed is an Erlenmeyer flask containing aqueous sodium hydroxide
H2SO4 (aq) NaOH (aq)  Na2SO4 (aq) H2O (l)
Follow the same method as before!

31. Molecular, Ionic and Net Ionic Equations
Complete Ionic Equations = CIE
Example #1
Sulfuric acid is placed is an Erlenmeyer flask containing aqueous sodium hydroxide
2 H+ (aq) SO42- (aq) Na+ (aq) OH- (aq) 
2 Na+ (aq) + SO42_ (aq) H2O (l)
Remember to depict: COEFFICIENT, CHARGE and STATE for each ion!
Spectator Ions = Na+ and SO42_

32. Molecular, Ionic and Net Ionic Equations
Net Ionic Equations = NIE
Example #1
Sulfuric acid is placed is an Erlenmeyer flask containing aqueous sodium hydroxide
Thus:
2 H+ (aq) OH− (aq)  2 H2O (l)
But the final equation must have LOWEST COEFFICIENT RATIO
Final Answer
H+ (aq) + OH_ (aq)  H2O (l)

33. Molecular, Ionic and Net Ionic Equations
Molecular Equations
Example #2
Solutions of plumbous nitrate and lithium bromide are mixed in a test tube
Pb(NO3)2 (aq) LiBr (aq)  PbBr2 (s) LiNO3 (aq)
Remember that a precipitant does NOT dissolve in water, therefore it DOES NOT DISSOCIATE.

34. Molecular, Ionic and Net Ionic Equations
Complete Ionic Equations
Example #2
Solutions of plumbous nitrate and lithium bromide are mixed in a test tube
Pb+ (aq) + 2 NO3− (aq) Li Br− (aq) 
PbBr2 (s) Li+ (aq) NO3− (aq)
Spectator Ions = Li+ and NO3−

35. Molecular, Ionic and Net Ionic Equations
Example #2
Solutions of plumbous nitrate and lithium bromide are mixed in a test tube
Pb+ (aq) Br− (aq)  PbBr2 (s)
Final Answer
Pb+ (aq) Br−  PbBr2 (s)

37. Single Displacement Reactions and Oxidation NumbersBy
Professor Walker Waugh

38. Single Displacement Reactions
One type of Oxidation Reduction (Redox) Reactions
Occurs via a Transfer of Electrons and you must assign the OXIDATION NUMBERS to each element. But 1st let’s discuss single displacement reactions.
General Equation:
A + BC  B + AC
Metal
Look at Activity Series

39. Single Displacement Reactions
ACTIVITY SERIES FOR METALS (and HYDROGEN)
Li highest activity  K Ca
Na If A is higher than B then YES there will be a reaction Mg Al
Zn  Zn+2
Cr  Cr+3
Fe  Fe+2 
Cd  Cd+2
Ni  Ni+2
Sn  Sn+2
Pb  Pb+2 H2
Cu  Cu If A is NOT higher than B then NO REACTION
Ag  Ag+1
Hg  Hg+2
Au  Au lowest activity

40. Single Displacement Reactions
Example #1
A solution of silver nitrate is placed in a test tube containing copper metal
Cu (s) + 2 AgNO3 (aq)  Cu(NO3)2 (aq) Ag (s)
A B C
Method:
Look at the Activity Series to determine if a reaction will occur
Exchange A and B
Write the correct product formulas with states.

41. Single Displacement Reactions
Remember:
Metals DO NOT have a CHARGE
Metals are just electrically neutral atoms that share electrons
Metals become CATIONS upon reacting (see charge on Activity Series)

42. Single Displacement Reactions
Example #2
Magnesium metal is placed in a beaker full of hydrochloric acid
Mg (s) HCl (aq)  MgCl2 + H2 (g)
A B C
Now let’s learn to assign OXIDATION NUMBERS.

44. Assigning Oxidation NumbersBy
Professor Walker Waugh

45. Assigning Oxidation Numbers
Remember O I L R I G
Oxidation
Is a
Loss of Electrons
Reduction
Gain of Electrons

46. Assigning Oxidation Numbers
Assigning Oxidation Numbers (state) = a book keeping system to see where the electron are flowing
Method
Elemental forms = 0
Monatomic Ions = charge
Overall Oxidation Number for a Polyatomic ions = charge
Overall Oxidation Number of a compound = 0
Oxygen = -2, except in peroxide, O22-, = -1
Hydrogen in covalent compound = +1
Hydrogen in ionic compound = -1

47. Assigning Oxidation Numbers
Example 1
Individual Oxidation Number = IOxN (written above the element)
Overall Oxidation Number = OOxN (written below the element)
IOxN
Mg (s) HCl (aq)  MgCl2 + H2 (g)
p p+1 p p+ 12 p p+1
e e- 0 e e- 10 e e- 1
= =
OOxN

48. Assigning Oxidation Numbers
Individual Oxidation Number = IOxN (written above the element) ONLY the individual oxidation numbers are reported
Overall Oxidation Number = OOxN (written below the element) The overall oxidation number is the mathematical operation and should NOT be reported
IOxN
Mg (s) HCl (aq)  MgCl2 + H2 (g)
OOxN = =
Always to MULTIPLY by the subscript when going from the IOxN to the OOxN and
Always to DIVIDE by the subscript when going from the OOxN to the IOxN.

49. Assigning Oxidation Numbers
Example 2
Assign the oxidation number for carbon in the oxalate ion:
IOxN
C2O42_
OOxN = -2
Answer: The oxidation number for carbon in oxalate is +3.

50. Assigning Oxidation Numbers
Example 2 METHOD:
1. The Overall Oxidation Number, OOxN, for a polyatomic ion is equal to the charge.
IOxN
C2O42_
OOxN = -2
2. Assign oxidation numbers to oxygen. That is the only other rule that you can use at this time.
IOxN
OOxN = -2

51. Assigning Oxidation Numbers
3. Assign oxidation numbers to oxygen. That is the only other rule that you can use at this time.
IOxN
C2O42_
OOxN = -2
4. To determine the OVERALL oxidation number, OOxN for an atom, you must multiple by the subscript. Thus for oxygen: -2 x 4 = -8.
OOxN = -2

52. Assigning Oxidation Numbers
5. The OVERALL oxidation number must equal -2. Therefore the OOxN for carbon is +6, since +6 – 8 = -2.
IOxN
C2O42_
OOxN = -2
6. Finally, to determine the INDIVIDUAL Oxidation Number, IOxN, of carbon, DIVIDE by the OOxN of carbon by the subscript of carbon. Thus, +6 / 2 = +3.
IOxN
Answer: The oxidation number for carbon in oxalate is +3.