Answer :
Sure, let's go through the problem step-by-step:
### Part (a): Balancing the Equation
The given chemical reaction is:
[tex]$\text{Al}_2\text{O}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3$[/tex]
To balance the equation, we need to ensure that we have the same number of each type of atom on both sides of the equation.
1. Aluminum (Al) atoms: There are 2 Al atoms on the left side (in Al₂O₃) and 1 Al atom on the right side (in Al(OH)₃). Therefore, we need 2 Al(OH)₃ on the right side to balance the Al atoms:
[tex]$\text{Al}_2\text{O}_3 + \text{H}_2\text{O} \rightarrow 2\text{Al(OH)}_3$[/tex]
2. Oxygen (O) atoms: There are 3 O atoms in Al₂O₃ and 1 O atom in H₂O on the left side, totaling 4 O atoms. On the right side, each Al(OH)₃ molecule has 3 O atoms, so with 2 Al(OH)₃, we have 6 O atoms. Therefore, to balance the O atoms, we need 3 H₂O molecules:
[tex]$\text{Al}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{Al(OH)}_3$[/tex]
3. Hydrogen (H) atoms: There are 6 H atoms in 3 H₂O molecules on the left side. On the right side, each Al(OH)₃ molecule has 3 H atoms, so with 2 Al(OH)₃, we have 6 H atoms.
Now the equation is balanced:
[tex]$\text{Al}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{Al(OH)}_3$[/tex]
### Part (b): Calculating the Mass of [tex]\( \text{Al(OH)}_3 \)[/tex] Produced
Given:
- Mass of [tex]\( \text{Al}_2\text{O}_3 \)[/tex]: 408 g
#### Step 1: Calculate the Molar Masses
- Molar mass of [tex]\( \text{Al}_2\text{O}_3 \)[/tex]:
- Aluminum (Al): 26.98 g/mol
- Oxygen (O): 16.00 g/mol
[tex]\[ \text{Molar mass of } \text{Al}_2\text{O}_3 = 2 \times 26.98 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol} = 101.96 \, \text{g/mol} \][/tex]
- Molar mass of [tex]\( \text{Al(OH)}_3 \)[/tex]:
- Aluminum (Al): 26.98 g/mol
- Oxygen (O): 16.00 g/mol
- Hydrogen (H): 1.008 g/mol
[tex]\[ \text{Molar mass of } \text{Al(OH)}_3 = 26.98 \, \text{g/mol} + 3 \times (16.00 \, \text{g/mol} + 1.008 \, \text{g/mol}) = 78.004 \, \text{g/mol} \][/tex]
#### Step 2: Convert Mass of [tex]\( \text{Al}_2\text{O}_3 \)[/tex] to Moles
[tex]\[ \text{Moles of } \text{Al}_2\text{O}_3 = \frac{\text{Mass of } \text{Al}_2\text{O}_3}{\text{Molar mass of } \text{Al}_2\text{O}_3} = \frac{408 \, \text{g}}{101.96 \, \text{g/mol}} = 4.00157 \, \text{moles} \][/tex]
#### Step 3: Determine the Moles of [tex]\( \text{Al(OH)}_3 \)[/tex] Produced
From the balanced equation, 1 mole of [tex]\( \text{Al}_2\text{O}_3 \)[/tex] produces 2 moles of [tex]\( \text{Al(OH)}_3 \)[/tex].
[tex]\[ \text{Moles of } \text{Al(OH)}_3 = 2 \times 4.00157 \, \text{moles} = 8.00314 \, \text{moles} \][/tex]
#### Step 4: Convert Moles of [tex]\( \text{Al(OH)}_3 \)[/tex] to Mass
[tex]\[ \text{Mass of } \text{Al(OH)}_3 = \text{Moles of } \text{Al(OH)}_3 \times \text{Molar mass of } \text{Al(OH)}_3 \][/tex]
[tex]\[ \text{Mass of } \text{Al(OH)}_3 = 8.00314 \, \text{moles} \times 78.004 \, \text{g/mol} = 624.28 \, \text{g} \][/tex]
### Final Answer
The mass of [tex]\( \text{Al(OH)}_3 \)[/tex] produced from 408 g of [tex]\( \text{Al}_2\text{O}_3 \)[/tex] is 624.28 g.
### Part (a): Balancing the Equation
The given chemical reaction is:
[tex]$\text{Al}_2\text{O}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3$[/tex]
To balance the equation, we need to ensure that we have the same number of each type of atom on both sides of the equation.
1. Aluminum (Al) atoms: There are 2 Al atoms on the left side (in Al₂O₃) and 1 Al atom on the right side (in Al(OH)₃). Therefore, we need 2 Al(OH)₃ on the right side to balance the Al atoms:
[tex]$\text{Al}_2\text{O}_3 + \text{H}_2\text{O} \rightarrow 2\text{Al(OH)}_3$[/tex]
2. Oxygen (O) atoms: There are 3 O atoms in Al₂O₃ and 1 O atom in H₂O on the left side, totaling 4 O atoms. On the right side, each Al(OH)₃ molecule has 3 O atoms, so with 2 Al(OH)₃, we have 6 O atoms. Therefore, to balance the O atoms, we need 3 H₂O molecules:
[tex]$\text{Al}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{Al(OH)}_3$[/tex]
3. Hydrogen (H) atoms: There are 6 H atoms in 3 H₂O molecules on the left side. On the right side, each Al(OH)₃ molecule has 3 H atoms, so with 2 Al(OH)₃, we have 6 H atoms.
Now the equation is balanced:
[tex]$\text{Al}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{Al(OH)}_3$[/tex]
### Part (b): Calculating the Mass of [tex]\( \text{Al(OH)}_3 \)[/tex] Produced
Given:
- Mass of [tex]\( \text{Al}_2\text{O}_3 \)[/tex]: 408 g
#### Step 1: Calculate the Molar Masses
- Molar mass of [tex]\( \text{Al}_2\text{O}_3 \)[/tex]:
- Aluminum (Al): 26.98 g/mol
- Oxygen (O): 16.00 g/mol
[tex]\[ \text{Molar mass of } \text{Al}_2\text{O}_3 = 2 \times 26.98 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol} = 101.96 \, \text{g/mol} \][/tex]
- Molar mass of [tex]\( \text{Al(OH)}_3 \)[/tex]:
- Aluminum (Al): 26.98 g/mol
- Oxygen (O): 16.00 g/mol
- Hydrogen (H): 1.008 g/mol
[tex]\[ \text{Molar mass of } \text{Al(OH)}_3 = 26.98 \, \text{g/mol} + 3 \times (16.00 \, \text{g/mol} + 1.008 \, \text{g/mol}) = 78.004 \, \text{g/mol} \][/tex]
#### Step 2: Convert Mass of [tex]\( \text{Al}_2\text{O}_3 \)[/tex] to Moles
[tex]\[ \text{Moles of } \text{Al}_2\text{O}_3 = \frac{\text{Mass of } \text{Al}_2\text{O}_3}{\text{Molar mass of } \text{Al}_2\text{O}_3} = \frac{408 \, \text{g}}{101.96 \, \text{g/mol}} = 4.00157 \, \text{moles} \][/tex]
#### Step 3: Determine the Moles of [tex]\( \text{Al(OH)}_3 \)[/tex] Produced
From the balanced equation, 1 mole of [tex]\( \text{Al}_2\text{O}_3 \)[/tex] produces 2 moles of [tex]\( \text{Al(OH)}_3 \)[/tex].
[tex]\[ \text{Moles of } \text{Al(OH)}_3 = 2 \times 4.00157 \, \text{moles} = 8.00314 \, \text{moles} \][/tex]
#### Step 4: Convert Moles of [tex]\( \text{Al(OH)}_3 \)[/tex] to Mass
[tex]\[ \text{Mass of } \text{Al(OH)}_3 = \text{Moles of } \text{Al(OH)}_3 \times \text{Molar mass of } \text{Al(OH)}_3 \][/tex]
[tex]\[ \text{Mass of } \text{Al(OH)}_3 = 8.00314 \, \text{moles} \times 78.004 \, \text{g/mol} = 624.28 \, \text{g} \][/tex]
### Final Answer
The mass of [tex]\( \text{Al(OH)}_3 \)[/tex] produced from 408 g of [tex]\( \text{Al}_2\text{O}_3 \)[/tex] is 624.28 g.
