In an experiment, `6.67 g` of `AlCl_(3)` was produced and `0.654 g` Al remainded unreacted. How many `g` atoms of `Al` and `Cl_(2)` were taken originally `(Al = 27, Cl = 35.5)`?

Options

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

Explanation

To solve the problem step by step, we will follow the given information and perform the necessary calculations.

### Step 1: Calculate the moles of AlCl₃ produced
We know that:
- Mass of AlCl₃ produced = 6.67 g
- Molar mass of AlCl₃ = Al (27 g/mol) + 3 × Cl (35.5 g/mol) = 27 + 106.5 = 133.5 g/mol

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

### Step 2: Calculate the moles of unreacted Al
We know that:
- Mass of unreacted Al = 0.654 g
- Molar mass of Al = 27 g/mol

Using the formula for moles:
\[ \text{Moles of unreacted Al} = \frac{0.654 \, \text{g}}{27 \, \text{g/mol}} \approx 0.0242 \, \text{moles} \approx 0.02 \, \text{moles} \]

### Step 3: Calculate the total moles of Al taken
The total moles of Al taken originally is the sum of the moles of Al that reacted to form AlCl₃ and the moles of unreacted Al.
\[ \text{Total moles of Al} = \text{Moles of AlCl}_3 + \text{Moles of unreacted Al} = 0.05 + 0.02 = 0.07 \, \text{moles} \]

### Step 4: Calculate the moles of Cl₂ taken
From the stoichiometry of the reaction, 1 mole of AlCl₃ is produced from 3 moles of Cl₂. Therefore, the moles of Cl₂ needed for the formation of AlCl₃ can be calculated as:
\[ \text{Moles of Cl}_2 = 3 \times \text{Moles of AlCl}_3 = 3 \times 0.05 = 0.15 \, \text{moles} \]

### Final Results
- Total grams of Al taken originally = 0.07 moles × 27 g/mol = 1.89 g
- Total grams of Cl₂ taken originally = 0.15 moles × 70.9 g/mol (molar mass of Cl₂) = 10.65 g

### Summary
- Original mass of Al taken = 1.89 g
- Original mass of Cl₂ taken = 10.65 g