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Metals - Extraction of metals (Iron and Aluminum)

Grade 12IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The method used to extract a metal depends on its position in the reactivity series. Metals above carbon, such as Aluminum (AlAl), are extracted using electrolysis. Metals below carbon, such as Iron (FeFe), are extracted by reduction using carbon or carbon monoxide.

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Iron is extracted in a Blast Furnace using Hematite (Fe2O3Fe_2O_3), Coke (CC), Limestone (CaCO3CaCO_3), and hot air.

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The primary reducing agent in the Blast Furnace is Carbon Monoxide (COCO), which is formed when Carbon Dioxide (CO2CO_2) reacts with excess Coke (CC).

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Limestone (CaCO3CaCO_3) is used to remove sandy impurities (SiO2SiO_2). It decomposes to Calcium Oxide (CaOCaO), which reacts with SiO2SiO_2 to form molten slag (CaSiO3CaSiO_3).

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Aluminum is extracted from Bauxite ore, which is purified to Alumina (Al2O3Al_2O_3). Because Al2O3Al_2O_3 has a very high melting point (approx. 2000∘C2000^\circ C), it is dissolved in molten Cryolite (Na3AlF6Na_3AlF_6) to lower the melting point to about 950∘C950^\circ C and improve conductivity.

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In the Hall-Héroult process for Aluminum, the Graphite (Carbon) anodes must be replaced periodically because the Oxygen gas (O2O_2) produced reacts with the carbon to form Carbon Dioxide (CO2CO_2).

📐Formulae

C(s)+O2(g)→CO2(g)C(s) + O_2(g) \rightarrow CO_2(g) (Combustion of Coke)

CO2(g)+C(s)→2CO(g)CO_2(g) + C(s) \rightarrow 2CO(g) (Formation of Reducing Agent)

Fe2O3(s)+3CO(g)→2Fe(l)+3CO2(g)Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(l) + 3CO_2(g) (Reduction of Iron Ore)

CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g) (Thermal Decomposition of Limestone)

CaO(s)+SiO2(s)→CaSiO3(l)CaO(s) + SiO_2(s) \rightarrow CaSiO_3(l) (Slag Formation)

Al3++3e−→Al(l)Al^{3+} + 3e^- \rightarrow Al(l) (Reduction at Cathode)

2O2−→O2(g)+4e−2O^{2-} \rightarrow O_2(g) + 4e^- (Oxidation at Anode)

C(s)+O2(g)→CO2(g)C(s) + O_2(g) \rightarrow CO_2(g) (Anode Erosion)

💡Examples

Problem 1:

In the extraction of Iron, calculate the mass of Iron produced from 160160 tonnes of Hematite (Fe2O3Fe_2O_3), assuming the ore is 100%100\% pure and the reaction goes to completion.

Solution:

  1. Determine Molar Masses: Mr(Fe2O3)=(2×56)+(3×16)=160M_r(Fe_2O_3) = (2 \times 56) + (3 \times 16) = 160 g/mol. Ar(Fe)=56A_r(Fe) = 56 g/mol.
  2. Use the stoichiometry from the equation: Fe2O3+3CO→2Fe+3CO2Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2.
  3. Ratio is 11 mole Fe2O3:2Fe_2O_3 : 2 moles FeFe.
  4. Mass of Fe=2×56160×160 tonnes=112 tonnesFe = \frac{2 \times 56}{160} \times 160\text{ tonnes} = 112\text{ tonnes}.

Explanation:

The balanced chemical equation shows that 11 mole of Hematite yields 22 moles of Iron. Since the molar mass of Hematite is 160160 and we have 160160 tonnes, we produce 2×562 \times 56 tonnes of Iron.

Problem 2:

Explain the role of Cryolite in the electrolysis of Aluminum Oxide (Al2O3Al_2O_3).

Solution:

Pure Alumina has a melting point of over 2000∘C2000^\circ C. Dissolving it in molten Cryolite (Na3AlF6Na_3AlF_6) allows the process to happen at 950∘C950^\circ C.

Explanation:

Cryolite acts as a solvent that lowers the operating temperature, significantly reducing energy costs and preventing the need for materials that can withstand extreme heat. It also provides better electrical conductivity for the electrolyte.

Extraction of metals (Iron and Aluminum) Revision - Grade 12 Chemistry IGCSE