Q-Alum
Prem SattsangiCopyright 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:
NH4Al(SO4)2.12H2O
LiAl(SO4)2.12H2O
KAl(SO4)2.12H2O
Ammonium alum
Lithium alum
Common alum
#2 What is the formula of?
Ammonium ion
Sulfate ion SO42-
Potassium Hydroxide KOH
Aluminum hydroxideAl(OH)3
Potassium aluminum hydroxide
KAl(OH)4
Potassium aluminum sulfate KAl(SO4)2
NH4+
Ammonium sulfate (NH4)2SO4
#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
(Hint: 3H+ + 3e- 3H =1.5H2)
Al(s)+KOH(aq)+3H2O(l) KAl(OH)4(aq) + 1.5H2(g)
Check: 2Al (getting oxidized) 2Al3+ + 6e-
6H+ + 6e- (getting reduced) 3H2
2Al(s)+2KOH(aq)+ 6H2O(l) 2KAl(OH)4(aq) + 3H2(g)
(1 Al Al3+ + 3e-)
#4 Amphoteric Aluminum hydroxideReacts with acid as well as with base.
Most Hydroxides are: Basic, React with acid
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)
Al(OH)3(aq) is Basic as well as Acidic.
It is AMPHOTERIC.
#5 Complete the Following Equations
2Al(OH)3(s) + 3H2SO4(aq) + Al2(SO4)3(aq) + 6HOH (l)
Al2(SO4)3(aq) + K2SO4(aq) + 24H2O 2KAl(SO4)2.12H2O
Al(OH)3(aq) + KOH(aq) KAl(OH)4(aq)
2Al(s) + 2KOH(aq) + 6H2O 2KAl(OH)4 (aq) + 3H2
2KAl(OH)4(aq) + H2SO4(aq) 2Al(OH)3(s) + K2SO4(aq)
#6 Theoretical and %yield (p.107) Given the following information:
Aluminum to AlumFormula Al KAl(SO4)2.12H2O
FW 27 474Mass of Aluminum = 0.0345 gEXPERIMENTAL YIELD OF ALUM = 0.4159 g
= 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
= 68.7%
Calculate Theoretical yield of alum:
Calculate %Yield: = 0.4159 g x 100 0.6056
#7 Electrical Energy from Aluminum.
2Al + 3O2 Al2O3 + 1670 kJ [3.6x103 kJ = 1 kwh]
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