Q-Alum
Prem Sattsangi
Copyright 2007
Corrosion of metals
(Reaction of a metal with Oxygen)
• Aluminum and Iron, both get oxidized to form
Cations with +3 charge.
• Al  Al3+ + 3e• Fe  Fe3+ + 3e• With oxygen, they react to form similar oxides:
• 4Al(s) + 3O2(g)  2Al2O3(s) Forms a THIN
protective coating and prevents further oxidation.
• 4Fe(s) + 3O2(g)  2Fe2O3(s) (known as RUST)
• Rust is porous. Iron continues to rust.
#1 Alum
• Explain the term alum
• A Complex salt formed by combination of aluminum
sulfate and a Gr. IA metal or ammonium sulfate.
• Write the formula of:
Ammonium alum
NH4Al(SO4)2.12H2O
Lithium alum
LiAl(SO4)2.12H2O
Common alum
KAl(SO4)2.12H2O
#2 What is the formula of?
Ammonium ion
NH4+
Sulfate ion
SO42-
Ammonium sulfate
(NH4)2SO4
Potassium Hydroxide
KOH
Aluminum hydroxide
Al(OH)3
Potassium aluminum
hydroxide
KAl(OH)4
Potassium aluminum
sulfate
KAl(SO4)2
#3 Write The Redox Reaction when
Al(s) reacts with KOH(aq).
Al(s)+ KOH(aq) + H+OH- 
Ox.# = 0
= +1
KAl(OH)4(aq) + H2(g)
Al3+
Ox.# = 0
(1 Al  Al3+ + 3e-)
Al(s)+KOH(aq)+3H2O(l)  KAl(OH)4(aq) + 1.5H2(g)
(Hint: 3H+ + 3e-  3H =1.5H2)
2Al(s)+2KOH(aq)+ 6H2O(l)  2KAl(OH)4(aq) + 3H2(g)
Check: 2Al (getting oxidized)  2Al3+ + 6e6H+ + 6e- (getting reduced)  3H2
#4 Amphoteric Aluminum hydroxide
Reacts with acid as well as with base.
Most Hydroxides are:
Basic,
Fe(OH)3(s) + KOH(aq) 
No reaction
Fe(OH)3(s) + 3HCl(aq) 
FeCl3(aq) + 3H-OH(l)
Al(OH)3(aq) + 3HCl(aq) 
AlCl3(aq) + 3H-OH(l)
Al(OH)3(aq) + KOH(aq) 
K+(aq) + [Al(OH)4]-(aq)
React with
Al(OH)3(aq) is Basic as well as Acidic.
It is AMPHOTERIC.
acid
#5 Complete the Following Equations
2Al(s) + 2KOH(aq) + 6H2O 
Al(OH)3(aq) + KOH(aq) 
2KAl(OH)4(aq) + H2SO4(aq) 
2Al(OH)3(s) + 3H2SO4(aq) + D
2KAl(OH)4 (aq) + 3H2
KAl(OH)4(aq)
2Al(OH)3(s) + K2SO4(aq)
Al2(SO4)3(aq) + 6HOH (l)
Al2(SO4)3(aq) + K2SO4(aq) + 24H2O
2KAl(SO4)2.12H2O
#6 Theoretical and %yield (p.107)
Given the following information:
Aluminum
to
Alum
Formula
Al
KAl(SO4)2.12H2O
FW
27
474
Mass of Aluminum = 0.0345 g
EXPERIMENTAL YIELD OF ALUM = 0.4159 g
Calculate Theoretical yield of alum:
= 0.0345g Al x 1mol Al x 1 mol Alum x 474 g Alum
26.99 g Al 1 mol Al
1 mol Alum
= 0.6056 g Alum
Calculate %Yield:
= 0.4159 g x 100
0.6056
= 68.7%
#7 Electrical Energy from Aluminum.
2Al + 3O2  Al2O3 + 1670 kJ
kwh]
[3.6x103 kJ = 1
Aluminum 4.25g can be used to light a 0.025 kw bulb for
how many hours. USE Unit Conversion Factors to calculate.
4.25g Al x 1mol Al x 1mol Al2O3 x 1670 kJ
x 1kwh______
27g Al
2 mol Al
1mol Al2O3 3.6x103 kJ x 0.025 kw
= 1.5 h