`A 1.85 g` sample of mixture of `CuCl_(2)` and `CuBr_(2)` was dissolved in water and mixed thoroughly with `1.8 g ` portion of `AgCl`. After reaction, the solid which now dissoved contain `AgCl` and `AgBr` was filtered, dried and weighed to be `2.052 g.` What was the `%` by weight of `CuBr_(2)` in the mixture?

Explanation

To find the percentage by weight of `CuBr₂` in the mixture, we can follow these steps:

### Step 1: Write down the given data
- Mass of the mixture (CuCl₂ + CuBr₂) = 1.85 g
- Mass of AgCl = 1.8 g
- Mass of the solid after reaction (AgCl + AgBr) = 2.052 g

### Step 2: Define variables
Let:
- Mass of AgBr formed = x g
- Therefore, mass of AgCl formed = 2.052 g - x g

### Step 3: Calculate moles of AgCl and AgBr
Using the molar masses:
- Molar mass of AgCl = 143.5 g/mol
- Molar mass of AgBr = 188 g/mol

The moles of AgCl and AgBr can be expressed as:
- Moles of AgCl = (2.052 - x) / 143.5
- Moles of AgBr = x / 188

### Step 4: Apply conservation of moles
From the reaction:
- CuCl₂ reacts with AgCl to form AgCl
- CuBr₂ reacts with AgBr to form AgBr

The moles of CuCl₂ and CuBr₂ can be expressed as:
- Moles of CuCl₂ = Moles of AgCl
- Moles of CuBr₂ = 0.5 * Moles of AgBr

### Step 5: Set up the equation
Using the conservation of moles:
\[ \frac{2.052 - x}{143.5} = \frac{x}{188} \cdot 2 \]

### Step 6: Solve for x
Cross-multiplying gives:
\[ 188(2.052 - x) = 143.5(2x) \]
Expanding and rearranging:
\[ 376.416 - 188x = 287x \]
\[ 376.416 = 475x \]
\[ x = \frac{376.416}{475} \approx 0.791 g \]

### Step 7: Calculate the mass of CuBr₂
Since the mass of AgBr formed is x, we can find the moles of CuBr₂:
\[ \text{Moles of CuBr₂} = \frac{x}{188} \cdot 0.5 \]

Now, calculate the mass of CuBr₂:
\[ \text{Mass of CuBr₂} = \text{Moles of CuBr₂} \times 223.5 \]
Substituting the value of x:
\[ \text{Mass of CuBr₂} = \left(\frac{0.791}{188} \cdot 0.5\right) \times 223.5 \]
Calculating this gives:
\[ \text{Mass of CuBr₂} \approx 0.632 g \]

### Step 8: Calculate the percentage by weight of CuBr₂
Now, we can find the percentage by weight of CuBr₂ in the mixture:
\[ \text{Percentage of CuBr₂} = \left(\frac{\text{Mass of CuBr₂}}{\text{Total mass of mixture}}\right) \times 100 \]
Substituting the values:
\[ \text{Percentage of CuBr₂} = \left(\frac{0.632}{1.85}\right) \times 100 \approx 34.18\% \]

### Final Answer
The percentage by weight of `CuBr₂` in the mixture is approximately **34.18%**.

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