In an experiment, `6.67 g` of `AlCl_(3)` was produced and `0.54g` `Al` remained unreacted. How many moles of `Al` and `Cl_(2)` were taken originally?

Options

A. 0.070, 0.075
B. 0.07, 0.05
C. 0.02, 0.15
D. 0.02, 0.15

Explanation

To solve the problem step by step, we will follow the stoichiometric principles and calculations based on the information provided.

### Step 1: Determine the moles of unreacted Aluminum (Al)
Given:
- Mass of unreacted Aluminum (Al) = 0.54 g
- Molar mass of Aluminum (Al) = 27 g/mol

Using the formula for moles:
\[ \text{Moles of Al} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{0.54 \, \text{g}}{27 \, \text{g/mol}} = 0.02 \, \text{mol} \]

### Step 2: Determine the moles of Aluminum Chloride (AlCl₃) produced
Given:
- Mass of Aluminum Chloride (AlCl₃) produced = 6.67 g
- Molar mass of Aluminum Chloride (AlCl₃) = 133.5 g/mol

Using the formula for moles:
\[ \text{Moles of AlCl}_3 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{6.67 \, \text{g}}{133.5 \, \text{g/mol}} \approx 0.05 \, \text{mol} \]

### Step 3: Relate moles of AlCl₃ to moles of Chlorine (Cl₂)
From the balanced chemical equation:
\[ 2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3 \]
This indicates that 1 mole of AlCl₃ is produced from 3/2 moles of Cl₂. Therefore:
\[ \text{Moles of Cl}_2 = \frac{3}{2} \times \text{Moles of AlCl}_3 = \frac{3}{2} \times 0.05 \, \text{mol} = 0.075 \, \text{mol} \]

### Step 4: Calculate the initial moles of Aluminum (Al)
Let \(X\) be the initial moles of Aluminum. Since 0.02 moles of Aluminum remained unreacted and 0.05 moles were used to produce AlCl₃, we can set up the equation:
\[ X - 0.05 = 0.02 \]
Solving for \(X\):
\[ X = 0.02 + 0.05 = 0.07 \, \text{mol} \]

### Summary of Results
- Initial moles of Aluminum (Al) = 0.07 mol
- Initial moles of Chlorine (Cl₂) = 0.075 mol

### Final Answer
- Moles of Al taken originally = 0.07 mol
- Moles of Cl₂ taken originally = 0.075 mol

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