Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To determine the maximum mass of aluminum chloride (AlCl₃) that can be formed when reacting 22.0 g of aluminum (Al) with 27.0 g of chlorine gas (Cl₂), follow these steps:
### Step 1: Convert masses to moles
First, we need to calculate the moles of each reactant. The molar masses are:
- Aluminum (Al): 26.98 g/mol
- Chlorine gas (Cl₂): 70.90 g/mol
#### Moles of aluminum:
[tex]\[ \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} \][/tex]
[tex]\[ \text{moles of Al} = \frac{22.0 \, \text{g}}{26.98 \, \text{g/mol}} \approx 0.815 \, \text{mol} \][/tex]
#### Moles of chlorine gas:
[tex]\[ \text{moles of Cl₂} = \frac{\text{mass of Cl₂}}{\text{molar mass of Cl₂}} \][/tex]
[tex]\[ \text{moles of Cl₂} = \frac{27.0 \, \text{g}}{70.90 \, \text{g/mol}} \approx 0.381 \, \text{mol} \][/tex]
### Step 2: Determine the limiting reactant
The balanced chemical equation is:
[tex]\[ 2 Al (s) + 3 Cl_2 (g) \rightarrow 2 AlCl_3 (s) \][/tex]
From the stoichiometry of the reaction:
- 2 moles of Al react with 3 moles of Cl₂ to produce 2 moles of AlCl₃.
#### Moles of AlCl₃ from Al:
[tex]\[ \text{moles of AlCl₃ from Al} = \text{moles of Al} \][/tex]
[tex]\[ \text{moles of AlCl₃ from Al} \approx 0.815 \, \text{mol} \][/tex]
#### Moles of AlCl₃ from Cl₂:
[tex]\[ \text{moles of AlCl₃ from Cl₂} = \frac{\text{moles of Cl₂}}{3} \times 2 \][/tex]
[tex]\[ \text{moles of AlCl₃ from Cl₂} = \frac{0.381 \, \text{mol}}{3} \times 2 \approx 0.254 \, \text{mol} \][/tex]
The limiting reactant is the one that produces fewer moles of the product, which in this case is chlorine gas (Cl₂).
### Step 3: Calculate the mass of aluminum chloride produced
Using the moles of AlCl₃ determined by the limiting reactant:
[tex]\[ \text{moles of AlCl₃} \approx 0.254 \, \text{mol} \][/tex]
The molar mass of AlCl₃ is 133.34 g/mol.
[tex]\[ \text{mass of AlCl₃} = \text{moles of AlCl₃} \times \text{molar mass of AlCl₃} \][/tex]
[tex]\[ \text{mass of AlCl₃} \approx 0.254 \, \text{mol} \times 133.34 \, \text{g/mol} \approx 33.9 \, \text{g} \][/tex]
### Final Answer
The maximum mass of aluminum chloride (AlCl₃) that can be formed is approximately:
[tex]\[ 33.9 \, \text{g} \][/tex]
This answer is rounded to three significant figures, as requested.
### Step 1: Convert masses to moles
First, we need to calculate the moles of each reactant. The molar masses are:
- Aluminum (Al): 26.98 g/mol
- Chlorine gas (Cl₂): 70.90 g/mol
#### Moles of aluminum:
[tex]\[ \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} \][/tex]
[tex]\[ \text{moles of Al} = \frac{22.0 \, \text{g}}{26.98 \, \text{g/mol}} \approx 0.815 \, \text{mol} \][/tex]
#### Moles of chlorine gas:
[tex]\[ \text{moles of Cl₂} = \frac{\text{mass of Cl₂}}{\text{molar mass of Cl₂}} \][/tex]
[tex]\[ \text{moles of Cl₂} = \frac{27.0 \, \text{g}}{70.90 \, \text{g/mol}} \approx 0.381 \, \text{mol} \][/tex]
### Step 2: Determine the limiting reactant
The balanced chemical equation is:
[tex]\[ 2 Al (s) + 3 Cl_2 (g) \rightarrow 2 AlCl_3 (s) \][/tex]
From the stoichiometry of the reaction:
- 2 moles of Al react with 3 moles of Cl₂ to produce 2 moles of AlCl₃.
#### Moles of AlCl₃ from Al:
[tex]\[ \text{moles of AlCl₃ from Al} = \text{moles of Al} \][/tex]
[tex]\[ \text{moles of AlCl₃ from Al} \approx 0.815 \, \text{mol} \][/tex]
#### Moles of AlCl₃ from Cl₂:
[tex]\[ \text{moles of AlCl₃ from Cl₂} = \frac{\text{moles of Cl₂}}{3} \times 2 \][/tex]
[tex]\[ \text{moles of AlCl₃ from Cl₂} = \frac{0.381 \, \text{mol}}{3} \times 2 \approx 0.254 \, \text{mol} \][/tex]
The limiting reactant is the one that produces fewer moles of the product, which in this case is chlorine gas (Cl₂).
### Step 3: Calculate the mass of aluminum chloride produced
Using the moles of AlCl₃ determined by the limiting reactant:
[tex]\[ \text{moles of AlCl₃} \approx 0.254 \, \text{mol} \][/tex]
The molar mass of AlCl₃ is 133.34 g/mol.
[tex]\[ \text{mass of AlCl₃} = \text{moles of AlCl₃} \times \text{molar mass of AlCl₃} \][/tex]
[tex]\[ \text{mass of AlCl₃} \approx 0.254 \, \text{mol} \times 133.34 \, \text{g/mol} \approx 33.9 \, \text{g} \][/tex]
### Final Answer
The maximum mass of aluminum chloride (AlCl₃) that can be formed is approximately:
[tex]\[ 33.9 \, \text{g} \][/tex]
This answer is rounded to three significant figures, as requested.
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
