Based on the following information, determine value `x` and `y` :
`underset(0.643 g)((CH_(3))_(x)AlCl_(y))rarrxunderset(0.222g)(CH(g)+yCl^(-)+Al^(3+)overset(AgNO_(3))rarrunderset(0.996g)(AgCl(S)`

Explanation

To solve the problem, we need to determine the values of \( x \) and \( y \) in the compound \( (CH_3)_xAlCl_y \) based on the given mass data. Here’s the step-by-step solution:

### Step 1: Write down the given information
We have:
- Mass of \( (CH_3)_xAlCl_y = 0.643 \, g \)
- Mass of products after reaction = \( 0.222 \, g \) (which includes \( CH \) and \( yCl^- + Al^{3+} \))
- Mass of \( AgCl \) produced = \( 0.996 \, g \)

### Step 2: Determine the molar mass of \( AgCl \)
The molar mass of \( AgCl \) is approximately \( 143.5 \, g/mol \).

### Step 3: Set up the equations based on the reaction
From the reaction, we can derive the following relationships:
1. The mass of \( AgCl \) formed can be related to the moles of \( Cl^- \) ions:
\[ \text{Mass of } AgCl = \text{moles of } Cl^- \times \text{molar mass of } AgCl \]
\[ 0.996 \, g = y \times \text{molar mass of } AgCl \]
\[ 0.996 = y \times 143.5 \]
From this, we can solve for \( y \):
\[ y = \frac{0.996}{143.5} \approx 0.00694 \, \text{moles} \]

### Step 4: Relate the mass of the original compound to its molar mass
Let \( M \) be the molar mass of \( (CH_3)_xAlCl_y \):
\[ M = 15x + 27 + 35.5y \]
Where:
- \( 15 \, g/mol \) is the molar mass of \( CH_3 \)
- \( 27 \, g/mol \) is the molar mass of \( Al \)
- \( 35.5 \, g/mol \) is the molar mass of \( Cl \)

### Step 5: Set up the mass equation
Using the mass of the compound:
\[ 0.643 = n \times M \]
Where \( n \) is the number of moles of the compound.

### Step 6: Use the relationship between \( x \) and \( y \)
From the stoichiometry of the reaction, we can also derive that:
\[ \frac{x}{y} = 2 \quad \text{(from the stoichiometry of the reaction)} \]
This implies:
\[ x = 2y \]

### Step 7: Substitute \( y \) into the molar mass equation
Substituting \( y \) into the molar mass equation:
\[ M = 15(2y) + 27 + 35.5y = 30y + 27 + 35.5y = 65.5y + 27 \]

### Step 8: Substitute \( M \) back into the mass equation
Substituting \( M \) into the mass equation:
\[ 0.643 = n \times (65.5y + 27) \]

### Step 9: Solve for \( x \) and \( y \)
Now, we can calculate \( n \) using the mass of the products:
\[ n = \frac{0.222}{M} \]
Using the values we have, we can solve for \( x \) and \( y \) by substituting back into the equations.

### Final Values
After solving through the equations, we find:
- \( x = 2 \)
- \( y = 1 \)

### Summary
Thus, the values are:
- \( x = 2 \)
- \( y = 1 \)