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.

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Presentation transcript:

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  Al e - Fe  Fe e - With oxygen, they react to form similar oxides: 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s) Forms a THIN protective coating and prevents further oxidation. 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (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: NH 4 Al(SO 4 ) 2.12H 2 O LiAl(SO 4 ) 2.12H 2 O KAl(SO 4 ) 2.12H 2 O Ammonium alum Lithium alum Common alum

#2 What is the formula of? Ammonium ion Sulfate ion SO 4 2- Potassium Hydroxide KOH Aluminum hydroxide Al(OH) 3 Potassium aluminum hydroxide KAl(OH) 4 Potassium aluminum sulfate KAl(SO 4 ) 2 NH 4 + Ammonium sulfate(NH 4 ) 2 SO 4

#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) + H 2 (g) Al 3+ Ox.# = 0 (Hint: 3H + + 3e -  3H =1.5H 2 ) Al(s)+KOH(aq)+3H 2 O(l)  KAl(OH) 4 (aq) + 1.5H 2 (g) Check: 2Al (getting oxidized)  2Al e - 6H + + 6e - (getting reduced)  3H 2 2Al(s)+2KOH(aq)+ 6H 2 O(l)  2KAl(OH) 4 (aq) + 3H 2 (g) (1 Al  Al e - )

#4 Amphoteric Aluminum hydroxide Reacts 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)  FeCl 3 (aq) + 3H-OH(l) Al(OH) 3 (aq) + 3HCl(aq)  AlCl 3 (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) + 3H 2 SO 4 (aq) +   Al 2 (SO 4 ) 3 (aq) + 6HOH (l) Al 2 (SO 4 ) 3 (aq) + K 2 SO 4 (aq) + 24H 2 O  2KAl(SO 4 ) 2.12H 2 O Al(OH) 3 (aq) + KOH(aq)  KAl(OH) 4 (aq) 2Al(s) + 2KOH(aq) + 6H 2 O  2KAl(OH) 4 (aq) + 3H 2 2KAl(OH) 4 (aq) + H 2 SO 4 (aq)  2Al(OH) 3 (s) + K 2 SO 4 (aq)

#6 Theoretical and %yield (p.107) Given the following information: Aluminumto Alum FormulaAlKAl(SO 4 ) 2.12H 2 O FW27474 Mass of Aluminum = g EXPERIMENTAL YIELD OF ALUM = g =0.0345g Al x 1mol Al x 1 mol Alum x 474 g Alum g Al 1 mol Al 1 mol Alum = g Alum = 68.7% Calculate Theoretical yield of alum: Calculate %Yield: = g x

#7 Electrical Energy from Aluminum. 2Al + 3O 2  Al 2 O kJ [3.6x10 3 kJ = 1 kwh] Aluminum 4.25g can be used to light a kw bulb for how many hours. USE Unit Conversion Factors to calculate. 4.25g Al x 1mol Al x 1mol Al 2 O 3 x 1670 kJ x 1kwh______ 27g Al 2 mol Al 1mol Al 2 O 3 3.6x10 3 kJ x kw = 1.5 h