To determine if the reaction between aluminum (Al) and oxygen (O₂) to form aluminum oxide (Al₂O₃) is properly balanced, we first need to write the unbalanced chemical equation:
\[ \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \]
Next, we count the number of each type of atom on both sides of the equation:
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Reactants:
- Aluminum (Al): 1
- Oxygen (O): 2 (from O₂)
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Products:
- Aluminum (Al): 2 (from Al₂O₃)
- Oxygen (O): 3 (from Al₂O₃)
Now, let's analyze the counts:
- Aluminum: 1 on the reactant side and 2 on the product side (unbalanced)
- Oxygen: 2 on the reactant side and 3 on the product side (unbalanced)
To balance the equation, we try to make the number of both aluminum and oxygen atoms equal on both sides.
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To balance aluminum, we can multiply Al by 2 on the reactant side:
\[ 2\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \]
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Now count again:
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Reactants:
- Aluminum: 2
- Oxygen: 2 (from O₂)
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Products:
- Aluminum: 2 (from Al₂O₃)
- Oxygen: 3 (from Al₂O₃)
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Next, we need to balance the oxygen. Notice we have 2 on the reactant side and 3 on the product side. To balance the oxygen, we can multiply O₂ by \( \frac{3}{2} \):
\[ 2\text{Al} + \frac{3}{2}\text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \]
However, we typically avoid fractions in chemical equations, so we can multiply through by 2 to eliminate the fraction:
\[ 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \]
Final Counts:
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Reactants:
- Aluminum: 4 (from 4 Al)
- Oxygen: 6 (from 3 O₂)
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Products:
- Aluminum: 4 (from 2 Al₂O₃)
- Oxygen: 6 (from 2 Al₂O₃)
Now the equation is balanced.
Based on the above analysis, the initial equation was not balanced, and it needed adjustments to the coefficients of the reactants and products to achieve balance. Therefore, the correct option is A. This equation incorrectly increases the number of oxygen molecules on the reactant side, leading to an excess of oxygen atoms on the reactant side compared to the product side.
2 answers