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