Find solutions to your problems with the expert advice available on IDNLearn.com. Ask any question and get a thorough, accurate answer from our community of experienced professionals.
Sagot :
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.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Find precise solutions at IDNLearn.com. Thank you for trusting us with your queries, and we hope to see you again.
