Answer :
To balance the chemical equation:
[tex]\[ Al + O_2 \rightarrow Al_2O_3 \][/tex]
we need to ensure that the number of each type of atom on the reactants side (left side) is equal to the number on the products side (right side). Let's go through the process step by step:
1. Identify the elements involved:
The elements in this reaction are Aluminum (Al) and Oxygen (O).
2. Write the initial counts of atoms:
- On the left side (reactants):
- 1 Al (from Al)
- 2 O (from O_2)
- On the right side (products):
- 2 Al (from Al_2O_3)
- 3 O (from Al_2O_3)
3. Balance the number of Al atoms:
On the products side, there are 2 Aluminum atoms (Al_2O_3). To balance this, we need 2 Al on the reactants side:
[tex]\[ 2Al + O_2 \rightarrow Al_2O_3 \][/tex]
Now the counts are:
- Left side:
- 2 Al
- 2 O
- Right side:
- 2 Al
- 3 O
4. Balance the number of O atoms:
On the products side, there are 3 Oxygen atoms (Al_2O_3), but on the reactants side, there are only 2 Oxygen atoms (from O_2). Since we have molecules of \(O_2\), to balance the Oxygen atoms, we need \( \frac{3}{2}O_2\):
[tex]\[ 2Al + \frac{3}{2}O_2 \rightarrow Al_2O_3 \][/tex]
However, using fractional coefficients is not ideal in chemistry, so we will multiply all coefficients by 2 to get whole numbers:
[tex]\[ 2 \times 2Al + 2 \times \frac{3}{2}O_2 \rightarrow 2 \times Al_2O_3 \][/tex]
Simplifying, we get:
[tex]\[ 4Al + 3O_2 \rightarrow 2Al_2O_3 \][/tex]
5. Verify the balanced equation:
- Left side:
- 4 Al
- 3 \(O_2\) (which gives 6 O atoms)
- Right side:
- 4 Al (from 2 \(Al_2O_3\), since each \(Al_2O_3\) has 2 Al)
- 6 O (from 2 \(Al_2O_3\), since each \(Al_2O_3\) has 3 O)
Both sides now have 4 Al and 6 O atoms, so the equation is balanced.
The balanced equation is:
[tex]\[ 4Al + 3O_2 \rightarrow 2Al_2O_3 \][/tex]
[tex]\[ Al + O_2 \rightarrow Al_2O_3 \][/tex]
we need to ensure that the number of each type of atom on the reactants side (left side) is equal to the number on the products side (right side). Let's go through the process step by step:
1. Identify the elements involved:
The elements in this reaction are Aluminum (Al) and Oxygen (O).
2. Write the initial counts of atoms:
- On the left side (reactants):
- 1 Al (from Al)
- 2 O (from O_2)
- On the right side (products):
- 2 Al (from Al_2O_3)
- 3 O (from Al_2O_3)
3. Balance the number of Al atoms:
On the products side, there are 2 Aluminum atoms (Al_2O_3). To balance this, we need 2 Al on the reactants side:
[tex]\[ 2Al + O_2 \rightarrow Al_2O_3 \][/tex]
Now the counts are:
- Left side:
- 2 Al
- 2 O
- Right side:
- 2 Al
- 3 O
4. Balance the number of O atoms:
On the products side, there are 3 Oxygen atoms (Al_2O_3), but on the reactants side, there are only 2 Oxygen atoms (from O_2). Since we have molecules of \(O_2\), to balance the Oxygen atoms, we need \( \frac{3}{2}O_2\):
[tex]\[ 2Al + \frac{3}{2}O_2 \rightarrow Al_2O_3 \][/tex]
However, using fractional coefficients is not ideal in chemistry, so we will multiply all coefficients by 2 to get whole numbers:
[tex]\[ 2 \times 2Al + 2 \times \frac{3}{2}O_2 \rightarrow 2 \times Al_2O_3 \][/tex]
Simplifying, we get:
[tex]\[ 4Al + 3O_2 \rightarrow 2Al_2O_3 \][/tex]
5. Verify the balanced equation:
- Left side:
- 4 Al
- 3 \(O_2\) (which gives 6 O atoms)
- Right side:
- 4 Al (from 2 \(Al_2O_3\), since each \(Al_2O_3\) has 2 Al)
- 6 O (from 2 \(Al_2O_3\), since each \(Al_2O_3\) has 3 O)
Both sides now have 4 Al and 6 O atoms, so the equation is balanced.
The balanced equation is:
[tex]\[ 4Al + 3O_2 \rightarrow 2Al_2O_3 \][/tex]
