Metallurgy and Chemistry of the MainGroup Metals

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Metallurgy and Chemistry of the Main-Group Metals

• Chapter 21

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Metallurgy

• The scientific study of the production of metals
  from their ores and the making of alloys having
  various useful properties.

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Metal Production

• Preliminary Treatment
• Concentration or chemical modification to
  facilitate later steps.
• Reduction
• Metal ions are reduced to form the free metal by
  electrolysis or chemical means.
• Refining
• Purification of the free metal generated by
  electrolysis.

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Metallurgy and Stoichiometry

• How many kilograms of iron can be produced from
  1.00 kg of hydrogen, H2, when you reduce
  iron(III) oxide?

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Metallurgy and Stoichiometry

• How many kilograms of iron can be produced from
  1.00 kg of hydrogen, H2, when you reduce
  iron(III) oxide using hydrogen gas?

First balance the chemical equation,
Fe2O3 (s) H2 (g) Fe (s) H2O (g)
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Metallurgy and Stoichiometry

• How many kilograms of iron can be produced from
  1.00 kg of hydrogen, H2, when you reduce
  iron(III) oxide using hydrogen gas?

First balance the chemical equation,
Fe2O3 (s) 3 H2 (g) 2 Fe (s) 3 H2O (g)
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Metallurgy and Stoichiometry

• How many kilograms of iron can be produced from
  1.00 kg of hydrogen, H2, when you reduce
  iron(III) oxide using hydrogen gas?

Fe2O3 (s) 3 H2 (g) 2 Fe (s) 3 H2O (g)
Use the atomic masses and reaction stoichiometry
to solve,
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Metallurgy and Stoichiometry

• How many kilograms of iron can be produced from
  1.00 kg of hydrogen, H2, when you reduce
  iron(III) oxide using hydrogen gas?

Fe2O3 (s) 3 H2 (g) 2 Fe (s) 3 H2O (g)
Use the atomic masses and reaction stoichiometry
to solve,
1 mol H2 2.016 g H2
2 mol Fe 3 mol H2
55.85 g Fe 1 mol Fe
1.00 g H2 x
x
x
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Metallurgy and Stoichiometry

• How many kilograms of iron can be produced from
  1.00 kg of hydrogen, H2, when you reduce
  iron(III) oxide using hydrogen gas?

Fe2O3 (s) 3 H2 (g) 2 Fe (s) 3 H2O (g)
Use the atomic masses and reaction stoichiometry
to solve,
1 mol H2 2.016 g H2
2 mol Fe 3 mol H2
55.85 g Fe 1 mol Fe
1.00 g H2 x
x
x
1.85 x 104 g 18.5 kg Fe
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Superconductivity

• Ceramic (ionic solid) that shows no resistance to
  the flow of electric current below the critical
  temperature, Tc.

Resistance
Tc
Temperature
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Applications of High Tc Superconductors

• High Tc means greater than liquid N2 (77K)
• Electric wires
• Magnetic field maintenance
• High speed trains
• The Meissner effect
• Optical switching?

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Molecular Orbitals in Atomic Solids Conductivity
2 atomic orbitals
2 molecular orbitals (MO)
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Molecular Orbitals in Atomic Solids Conductivity
2 atomic orbitals
2 molecular orbitals (MO)
many MOs
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Molecular Orbitals in Atomic Solids Conductivity
Empty MOs
Filled MOs
2 atomic orbitals
2 molecular orbitals (MO)
many MOs
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Insulators
We can also consider different atoms or molecules
using this bonding.
Empty MOs
Empty MOs
Energy Gap
Filled MOs
Filled MOs
Insulator
Conductor
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Goldschmidt Process

• A method of preparing a metal by reduction of its
  oxide with powdered aluminum.

Cr2O3 (s) 2 Al (s) Al2O3 (l) 2 Cr (l)
?Ho -536 kJ
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Group IVa

• Lead(IV) oxide is a strong oxidizing agent. For
  example, lead(IV) oxide will oxidize hydrochloric
  acid to chlorine (Cl2). Write the balanced
  chemical equation for this process.

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Group IVa

• Lead(IV) oxide is a strong oxidizing agent. For
  example, lead(IV) oxide will oxidize hydrochloric
  acid to chlorine (Cl2). Write the balanced
  chemical equation for this process.

First write the two half reactions,
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Group IVa

• Lead(IV) oxide is a strong oxidizing agent. For
  example, lead(IV) oxide will oxidize hydrochloric
  acid to chlorine (Cl2). Write the balanced
  chemical equation for this process.

First write the two half reactions,
PbO2 4 H 2e- Pb2 2 H2O 2 Cl- Cl2
2 e-
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Group IVa

• Lead(IV) oxide is a strong oxidizing agent. For
  example, lead(IV) oxide will oxidize hydrochloric
  acid to chlorine (Cl2). Write the balanced
  chemical equation for this process.

First write the two half reactions,
PbO2 4 H 2e- Pb2 2 H2O 2 Cl- Cl2
2 e-
PbO2 4 H 2 Cl- Pb2 Cl2 2 H2O
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Group IVa

• Lead(IV) oxide is a strong oxidizing agent. For
  example, lead(IV) oxide will oxidize hydrochloric
  acid to chlorine (Cl2). Write the balanced
  chemical equation for this process.

PbO2 4 H 2e- Pb2 2 H2O 2 Cl- Cl2
2 e-
PbO2 4 HCl PbCl2 Cl2 2 H2O
With 2 Cl- added to both sides, the balanced
equation contains both HCl and PbCl2.