4Al (s)+ 3O2 (g) ->2Al2O3(s)
The Correct Answer and Explanation is :
The given chemical reaction is:
[ 4 \text{Al (s)} + 3 \text{O}_2 (g) \rightarrow 2 \text{Al}_2\text{O}_3 (s) ]
This represents the reaction between solid aluminum (Al) and oxygen gas (O₂) to form solid aluminum oxide (Al₂O₃). Here’s the breakdown and explanation:
Step 1: Understand the components
- Aluminum (Al) is a metal, commonly used in various applications due to its lightness, strength, and resistance to corrosion.
- Oxygen (O₂) is a diatomic molecule that supports combustion and various types of chemical reactions.
- Aluminum oxide (Al₂O₃) is a compound made up of aluminum and oxygen. It is a white, crystalline material and is commonly known as alumina. It forms a thin, protective oxide layer over aluminum metal, helping to protect it from further oxidation.
Step 2: Stoichiometry of the reaction
The balanced chemical equation indicates that:
- 4 moles of aluminum react with 3 moles of oxygen gas.
- This produces 2 moles of aluminum oxide.
Step 3: Reaction details
This reaction is an example of a combustion reaction where a metal reacts with oxygen to form an oxide. In this case, aluminum reacts with the oxygen in the air to form aluminum oxide. This process releases a significant amount of energy, often in the form of heat and light, and is commonly seen in the thermite reaction, where aluminum powder reacts with iron oxide (Fe₂O₃) to produce aluminum oxide and molten iron.
Step 4: Importance and applications
- Aluminum oxide (Al₂O₃) is used in various industries, including ceramics, as an abrasive, and as a material for producing aluminum metal through the Hall-Héroult process.
- The formation of an oxide layer on aluminum is what gives aluminum its resistance to corrosion, especially in outdoor environments.
Conclusion
This reaction showcases the combination of a metal (Al) with a non-metal (O₂) to form a simple ionic compound (Al₂O₃). It illustrates the principle of balancing chemical equations to ensure that atoms are conserved on both sides of the reaction.