Answer :
Sure, let's balance the given chemical equation step-by-step:
The unbalanced equation is:
[tex]\[ \text{H}_2\text{O} + \text{Al}_2\text{O}_3 \rightarrow \text{Al(OH)}_3 \][/tex]
Step 1: List the number of atoms of each element on both sides of the reaction.
Initially, the unbalanced equation looks like this:
[tex]\[ \text{H}_2\text{O} + \text{Al}_2\text{O}_3 \rightarrow \text{Al(OH)}_3 \][/tex]
- On the left-hand side (LHS):
- [tex]\( \text{H}_2\text{O} \)[/tex] has 2 Hydrogen (H) atoms and 1 Oxygen (O) atom.
- [tex]\( \text{Al}_2\text{O}_3 \)[/tex] has 2 Aluminum (Al) atoms and 3 Oxygen (O) atoms.
- On the right-hand side (RHS):
- [tex]\( \text{Al(OH)}_3 \)[/tex] has 1 Aluminum (Al) atom, 3 Oxygen (O) atoms, and 3 Hydrogen (H) atoms.
Step 2: Balance the Aluminum (Al) atoms.
Since there are 2 Aluminum (Al) atoms in [tex]\( \text{Al}_2\text{O}_3 \)[/tex], you need 2 [tex]\( \text{Al(OH)}_3 \)[/tex] units on the RHS to balance the Al atoms.
[tex]\[ H_2O + \text{Al}_2\text{O}_3 \rightarrow 2 \cdot \text{Al(OH)}_3 \][/tex]
Now, RHS has 2 Aluminum (Al) atoms, 6 Oxygen (O) atoms (from 2 OH groups), and 6 Hydrogen (H) atoms (from 2 OH groups).
Step 3: Balance the Hydrogen (H) atoms.
You've got 6 Hydrogen (H) atoms on the RHS. To balance those, you need 6 Hydrogen atoms on the LHS. Each molecule of [tex]\( \text{H}_2\text{O} \)[/tex] has 2 Hydrogen atoms, so you need 3 [tex]\( \text{H}_2\text{O} \)[/tex] molecules on the LHS.
[tex]\[ 3H_2O + \text{Al}_2\text{O}_3 \rightarrow 2 \cdot \text{Al(OH)}_3 \][/tex]
Step 4: Balance the Oxygen (O) atoms.
On the LHS, you have:
- 3 Oxygen (O) atoms from [tex]\( \text{H}_2\text{O} \)[/tex]
- 3 Oxygen (O) atoms from [tex]\( \text{Al}_2\text{O}_3 \)[/tex]
So, 3 + 3 = 6 Oxygen atoms in total.
On the RHS:
- You have 6 Oxygen (O) atoms from [tex]\( 2 \times \text{Al(OH)}_3 \)[/tex].
Everything is balanced:
[tex]\[ 3H_2O + \text{Al}_2\text{O}_3 \rightarrow 2 \cdot \text{Al(OH)}_3 \][/tex]
Step 5: Correct the mistake
We notice that by accounting for 6 oxygen atoms, we are still left with 6 H atoms from 3 H2O molecules and only 6 H from 2 Al(OH)3. It turns out we haven't arrived at an imbalanced scenario with wrong counts. So, we need to redo our counts.
Hence, the balanced equation should always be checked for counts:
- Oxygen atoms on LHS: (6H2O = 6O) + 3 (Al2O3)
- Oxygen on RHS: 9 from 6-OH groups.
Hence, it's concluded that by trial verification, the most accurate tally is:
[tex]\[ 6H_2O + \text{Al}_2\text{O}_3 \rightarrow 2 \cdot \text{Al(OH)}_3 \][/tex]
So the balanced chemical equation is:
[tex]\[ 6H_2O + Al_2O_3 \rightarrow 2Al(OH)_3 \][/tex]
This equation shows that 6 molecules of water react with 1 formula unit of aluminum oxide to form 2 formula units of aluminum hydroxide!
The unbalanced equation is:
[tex]\[ \text{H}_2\text{O} + \text{Al}_2\text{O}_3 \rightarrow \text{Al(OH)}_3 \][/tex]
Step 1: List the number of atoms of each element on both sides of the reaction.
Initially, the unbalanced equation looks like this:
[tex]\[ \text{H}_2\text{O} + \text{Al}_2\text{O}_3 \rightarrow \text{Al(OH)}_3 \][/tex]
- On the left-hand side (LHS):
- [tex]\( \text{H}_2\text{O} \)[/tex] has 2 Hydrogen (H) atoms and 1 Oxygen (O) atom.
- [tex]\( \text{Al}_2\text{O}_3 \)[/tex] has 2 Aluminum (Al) atoms and 3 Oxygen (O) atoms.
- On the right-hand side (RHS):
- [tex]\( \text{Al(OH)}_3 \)[/tex] has 1 Aluminum (Al) atom, 3 Oxygen (O) atoms, and 3 Hydrogen (H) atoms.
Step 2: Balance the Aluminum (Al) atoms.
Since there are 2 Aluminum (Al) atoms in [tex]\( \text{Al}_2\text{O}_3 \)[/tex], you need 2 [tex]\( \text{Al(OH)}_3 \)[/tex] units on the RHS to balance the Al atoms.
[tex]\[ H_2O + \text{Al}_2\text{O}_3 \rightarrow 2 \cdot \text{Al(OH)}_3 \][/tex]
Now, RHS has 2 Aluminum (Al) atoms, 6 Oxygen (O) atoms (from 2 OH groups), and 6 Hydrogen (H) atoms (from 2 OH groups).
Step 3: Balance the Hydrogen (H) atoms.
You've got 6 Hydrogen (H) atoms on the RHS. To balance those, you need 6 Hydrogen atoms on the LHS. Each molecule of [tex]\( \text{H}_2\text{O} \)[/tex] has 2 Hydrogen atoms, so you need 3 [tex]\( \text{H}_2\text{O} \)[/tex] molecules on the LHS.
[tex]\[ 3H_2O + \text{Al}_2\text{O}_3 \rightarrow 2 \cdot \text{Al(OH)}_3 \][/tex]
Step 4: Balance the Oxygen (O) atoms.
On the LHS, you have:
- 3 Oxygen (O) atoms from [tex]\( \text{H}_2\text{O} \)[/tex]
- 3 Oxygen (O) atoms from [tex]\( \text{Al}_2\text{O}_3 \)[/tex]
So, 3 + 3 = 6 Oxygen atoms in total.
On the RHS:
- You have 6 Oxygen (O) atoms from [tex]\( 2 \times \text{Al(OH)}_3 \)[/tex].
Everything is balanced:
[tex]\[ 3H_2O + \text{Al}_2\text{O}_3 \rightarrow 2 \cdot \text{Al(OH)}_3 \][/tex]
Step 5: Correct the mistake
We notice that by accounting for 6 oxygen atoms, we are still left with 6 H atoms from 3 H2O molecules and only 6 H from 2 Al(OH)3. It turns out we haven't arrived at an imbalanced scenario with wrong counts. So, we need to redo our counts.
Hence, the balanced equation should always be checked for counts:
- Oxygen atoms on LHS: (6H2O = 6O) + 3 (Al2O3)
- Oxygen on RHS: 9 from 6-OH groups.
Hence, it's concluded that by trial verification, the most accurate tally is:
[tex]\[ 6H_2O + \text{Al}_2\text{O}_3 \rightarrow 2 \cdot \text{Al(OH)}_3 \][/tex]
So the balanced chemical equation is:
[tex]\[ 6H_2O + Al_2O_3 \rightarrow 2Al(OH)_3 \][/tex]
This equation shows that 6 molecules of water react with 1 formula unit of aluminum oxide to form 2 formula units of aluminum hydroxide!
