To balance the chemical equation for the reaction of aluminum (Al) with oxygen (O₂) to form aluminum oxide (Al₂O₃), we need to ensure that the number of atoms of each element on the reactant side is equal to the number on the product side.
The unbalanced equation is:
\[ \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \]
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Count the number of atoms for each element:
- On the left side (reactants), we have:
- Aluminum (Al): 1
- Oxygen (O): 2 (from O₂)
- On the right side (products), we have:
- Aluminum (Al): 2 (from Al₂O₃)
- Oxygen (O): 3 (from Al₂O₃)
- On the left side (reactants), we have:
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Balance the Aluminum: We need 2 Al atoms on the left side to match the 2 Al atoms on the right side. We can do this by putting a coefficient of 2 in front of Al:
\[ 2 \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \]
- Balance the Oxygen:
- Now on the left side we still have O₂ (which provides 2 O atoms), and on the right side we have 3 O atoms in Al₂O₃. To balance the oxygen, we need to find a common multiple.
- The least common multiple of 2 (from O₂) and 3 (from Al₂O₃) is 6, so we will need 3 O₂ on the reactant side to give us 6 O atoms, and we will need 2 Al₂O₃ on the product side to give us 6 O atoms as well.
Putting this all together, the coefficients we need are:
\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \]
Therefore:
- Blank 1: 4
- Blank 2: 3
- Blank 3: 2
So the balanced equation is:
\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \]
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