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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.
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