Answer :
To balance the chemical equation for the oxidation of aluminum, the goal is to ensure that the number of each type of atom on the reactant side equals the number of each type of atom on the product side.
The unbalanced equation is:
[tex]\[ \_ \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \][/tex]
Let's use the following steps to balance this equation:
1. Identify the number of each type of atom on both sides:
- Reactants:
- Aluminum (Al): [tex]\( x \)[/tex] (since its coefficient is yet to be determined)
- Oxygen (O): 2 (from [tex]\(\text{O}_2\)[/tex])
- Products:
- Aluminum (Al): 2 (from [tex]\(\text{Al}_2\text{O}_3\)[/tex])
- Oxygen (O): 3 (from [tex]\(\text{Al}_2\text{O}_3\)[/tex])
2. Balance Aluminum (Al):
- The product side has 2 Al atoms in one [tex]\(\text{Al}_2\text{O}_3\)[/tex] molecule.
- To balance aluminum, set the coefficient of Al on the reactant side to 4:
[tex]\[ 4 \text{Al} + \_ \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
Now, we have 4 Al atoms on the reactant side and 4 Al atoms on the product side.
3. Balance Oxygen (O):
- The product side has 3 O atoms in one [tex]\(\text{Al}_2\text{O}_3\)[/tex] molecule.
- With the coefficient of 2 for [tex]\(\text{Al}_2\text{O}_3\)[/tex], we have [tex]\(2 \times 3 = 6\)[/tex] O atoms on the product side.
- To balance oxygen, set the coefficient of [tex]\(\text{O}_2\)[/tex] on the reactant side to 3:
[tex]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
Now, we have 6 O atoms on the reactant side (since [tex]\(3 \times 2 = 6\)[/tex]) and 6 O atoms on the product side.
4. Verify Balance:
- Aluminum (Al): 4 atoms on the reactant side and 4 atoms on the product side.
- Oxygen (O): 6 atoms on the reactant side and 6 atoms on the product side.
Thus, the balanced equation is:
[tex]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
The sequence of coefficients that should be placed in the blanks to balance the equation is:
[tex]\[ \boxed{4, 3, 2} \][/tex]
The unbalanced equation is:
[tex]\[ \_ \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \][/tex]
Let's use the following steps to balance this equation:
1. Identify the number of each type of atom on both sides:
- Reactants:
- Aluminum (Al): [tex]\( x \)[/tex] (since its coefficient is yet to be determined)
- Oxygen (O): 2 (from [tex]\(\text{O}_2\)[/tex])
- Products:
- Aluminum (Al): 2 (from [tex]\(\text{Al}_2\text{O}_3\)[/tex])
- Oxygen (O): 3 (from [tex]\(\text{Al}_2\text{O}_3\)[/tex])
2. Balance Aluminum (Al):
- The product side has 2 Al atoms in one [tex]\(\text{Al}_2\text{O}_3\)[/tex] molecule.
- To balance aluminum, set the coefficient of Al on the reactant side to 4:
[tex]\[ 4 \text{Al} + \_ \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
Now, we have 4 Al atoms on the reactant side and 4 Al atoms on the product side.
3. Balance Oxygen (O):
- The product side has 3 O atoms in one [tex]\(\text{Al}_2\text{O}_3\)[/tex] molecule.
- With the coefficient of 2 for [tex]\(\text{Al}_2\text{O}_3\)[/tex], we have [tex]\(2 \times 3 = 6\)[/tex] O atoms on the product side.
- To balance oxygen, set the coefficient of [tex]\(\text{O}_2\)[/tex] on the reactant side to 3:
[tex]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
Now, we have 6 O atoms on the reactant side (since [tex]\(3 \times 2 = 6\)[/tex]) and 6 O atoms on the product side.
4. Verify Balance:
- Aluminum (Al): 4 atoms on the reactant side and 4 atoms on the product side.
- Oxygen (O): 6 atoms on the reactant side and 6 atoms on the product side.
Thus, the balanced equation is:
[tex]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
The sequence of coefficients that should be placed in the blanks to balance the equation is:
[tex]\[ \boxed{4, 3, 2} \][/tex]
