Question
a. Balance the equation Al(s)+Cl2(g)=AlCl3(s)
b. If you begin with 3.2 g of aluminum and 5.4 g of chlorine, which is the limiting reactant?
c. How many grams of aluminum Chloride can be produced from the amount of limiting reactant available?
d. Calculate the mass of excess reactant that remains after te reaction is complete.
b. If you begin with 3.2 g of aluminum and 5.4 g of chlorine, which is the limiting reactant?
c. How many grams of aluminum Chloride can be produced from the amount of limiting reactant available?
d. Calculate the mass of excess reactant that remains after te reaction is complete.
Answers
GPT-4o mini
a. **Balancing the equation:**
The unbalanced chemical equation is:
\[ \text{Al(s)} + \text{Cl}_2(g) \rightarrow \text{AlCl}_3(s) \]
To balance the equation:
1. Each formula unit of aluminum chloride (AlCl₃) contains one aluminum atom and three chlorine atoms.
2. Therefore, for every one AlCl₃ produced, we need one Al and 1.5 Cl₂ molecules.
The balanced equation is:
\[ 2 \text{Al(s)} + 3 \text{Cl}_2(g) \rightarrow 2 \text{AlCl}_3(s) \]
b. **Determining the limiting reactant:**
1. Determine the molar masses:
- Molar mass of Al = 26.98 g/mol
- Molar mass of Cl = 35.45 g/mol
- Molar mass of AlCl₃ = 26.98 + (3 × 35.45) = 133.33 g/mol
2. Convert grams to moles:
- Moles of Al:
\[
\text{moles of Al} = \frac{3.2 \text{ g Al}}{26.98 \text{ g/mol}} \approx 0.119 moles
\]
- Moles of Cl₂:
\[
\text{moles of Cl₂} = \frac{5.4 \text{ g Cl}}{70.90 \text{ g/mol}} \approx 0.076 moles \text{ of Cl}_2
\]
3. From the balanced equation, we know the mole ratio is:
\[
2 \, \text{mol Al} \text{ : } 3 \, \text{mol Cl}_2
\]
For 0.119 moles of Al, the moles of Cl₂ required:
\[
\text{Required Cl}_2 = \frac{3}{2} \times 0.119 = 0.1785 \text{ moles}
\]
Since we only have 0.076 moles of Cl₂, Cl₂ is the limiting reactant.
c. **Calculating the grams of aluminum chloride produced:**
1. Using the limiting reactant Cl₂, we can find how much AlCl₃ is produced:
- According to the balanced equation, 3 moles of Cl₂ produce 2 moles of AlCl₃.
- Moles of AlCl₃ produced from 0.076 moles of Cl₂:
\[
\text{Moles of AlCl}_3 = \frac{2}{3} \times 0.076 \approx 0.0507 \text{ moles}
\]
2. Convert moles of AlCl₃ to grams:
\[
\text{Mass of AlCl}_3 = 0.0507 \text{ mol} \times 133.33 \text{ g/mol} \approx 6.76 \text{ g}
\]
d. **Calculating the mass of excess reactant that remains after the reaction is complete:**
1. Calculate how much aluminum is consumed:
- From the balanced equation, 2 moles of Al react with 3 moles of Cl₂.
- Moles of Al needed for 0.076 moles of Cl₂:
\[
\text{Moles of Al used} = \frac{2}{3} \times 0.076 \approx 0.0507 \text{ moles}
\]
2. Calculate the grams of aluminum used:
\[
\text{Mass of Al used} = 0.0507 \text{ mol} \times 26.98 \text{ g/mol} \approx 1.37 \text{ g}
\]
3. Calculate remaining aluminum:
\[
\text{Initial Al} = 3.2 \text{ g}, \quad \text{Mass of Al remaining} = 3.2 \text{ g} - 1.37 \text{ g} \approx 1.83 \text{ g}
\]
**Summary:**
- a. Balanced reaction: \( 2 \text{Al(s)} + 3 \text{Cl}_2(g) \rightarrow 2 \text{AlCl}_3(s) \)
- b. Limiting reactant: Cl₂
- c. Mass of aluminum chloride produced: \( \approx 6.76 \text{ g} \)
- d. Mass of excess aluminum remaining: \( \approx 1.83 \text{ g} \)
The unbalanced chemical equation is:
\[ \text{Al(s)} + \text{Cl}_2(g) \rightarrow \text{AlCl}_3(s) \]
To balance the equation:
1. Each formula unit of aluminum chloride (AlCl₃) contains one aluminum atom and three chlorine atoms.
2. Therefore, for every one AlCl₃ produced, we need one Al and 1.5 Cl₂ molecules.
The balanced equation is:
\[ 2 \text{Al(s)} + 3 \text{Cl}_2(g) \rightarrow 2 \text{AlCl}_3(s) \]
b. **Determining the limiting reactant:**
1. Determine the molar masses:
- Molar mass of Al = 26.98 g/mol
- Molar mass of Cl = 35.45 g/mol
- Molar mass of AlCl₃ = 26.98 + (3 × 35.45) = 133.33 g/mol
2. Convert grams to moles:
- Moles of Al:
\[
\text{moles of Al} = \frac{3.2 \text{ g Al}}{26.98 \text{ g/mol}} \approx 0.119 moles
\]
- Moles of Cl₂:
\[
\text{moles of Cl₂} = \frac{5.4 \text{ g Cl}}{70.90 \text{ g/mol}} \approx 0.076 moles \text{ of Cl}_2
\]
3. From the balanced equation, we know the mole ratio is:
\[
2 \, \text{mol Al} \text{ : } 3 \, \text{mol Cl}_2
\]
For 0.119 moles of Al, the moles of Cl₂ required:
\[
\text{Required Cl}_2 = \frac{3}{2} \times 0.119 = 0.1785 \text{ moles}
\]
Since we only have 0.076 moles of Cl₂, Cl₂ is the limiting reactant.
c. **Calculating the grams of aluminum chloride produced:**
1. Using the limiting reactant Cl₂, we can find how much AlCl₃ is produced:
- According to the balanced equation, 3 moles of Cl₂ produce 2 moles of AlCl₃.
- Moles of AlCl₃ produced from 0.076 moles of Cl₂:
\[
\text{Moles of AlCl}_3 = \frac{2}{3} \times 0.076 \approx 0.0507 \text{ moles}
\]
2. Convert moles of AlCl₃ to grams:
\[
\text{Mass of AlCl}_3 = 0.0507 \text{ mol} \times 133.33 \text{ g/mol} \approx 6.76 \text{ g}
\]
d. **Calculating the mass of excess reactant that remains after the reaction is complete:**
1. Calculate how much aluminum is consumed:
- From the balanced equation, 2 moles of Al react with 3 moles of Cl₂.
- Moles of Al needed for 0.076 moles of Cl₂:
\[
\text{Moles of Al used} = \frac{2}{3} \times 0.076 \approx 0.0507 \text{ moles}
\]
2. Calculate the grams of aluminum used:
\[
\text{Mass of Al used} = 0.0507 \text{ mol} \times 26.98 \text{ g/mol} \approx 1.37 \text{ g}
\]
3. Calculate remaining aluminum:
\[
\text{Initial Al} = 3.2 \text{ g}, \quad \text{Mass of Al remaining} = 3.2 \text{ g} - 1.37 \text{ g} \approx 1.83 \text{ g}
\]
**Summary:**
- a. Balanced reaction: \( 2 \text{Al(s)} + 3 \text{Cl}_2(g) \rightarrow 2 \text{AlCl}_3(s) \)
- b. Limiting reactant: Cl₂
- c. Mass of aluminum chloride produced: \( \approx 6.76 \text{ g} \)
- d. Mass of excess aluminum remaining: \( \approx 1.83 \text{ g} \)