A 2 lit solution contains 0.04 mol of each of `[CO(NH_(3))_(5)SO_(4)]Br` and `[CO(NH_(3))_(5)Br]SO_(4)`. To 1 lit of this solution, excess of `AgNO_(3)` is added. To the remaining solution of excess of `BaCl_(2)` is added. The amounts of precipitated salts, respectively, are

Options

A. 0.01 mol & 0.01 mol
B. 0.01 mol & 0.02 mol
C. 0.02 mol & 0.01 mol
D. 0.02 mol & 0.02 mol

Answer

To solve the problem step by step, we will analyze the situation involving the two complexes and the reactions with silver nitrate and barium chloride.

### Step 1: Determine the moles of each complex in the 1-liter solution
Given that the total volume of the solution is 2 liters and contains 0.04 moles of each complex, when we take 1 liter of this solution, the moles of each complex will be halved.

**Calculation:**
- Moles of each complex in 1 liter = \( \frac{0.04 \, \text{mol}}{2} = 0.02 \, \text{mol} \)

### Step 2: Identify the ions present in the complexes
The two complexes are:
1. \([Co(NH_3)_5SO_4]Br\)
2. \([Co(NH_3)_5Br]SO_4\)

In these complexes:
- The bromide ion (Br\(^-\)) is outside the coordination sphere in the first complex.
- The sulfate ion (SO\(_4^{2-}\)) is outside the coordination sphere in the second complex.

### Step 3: Reaction with excess AgNO\(_3\)
When excess AgNO\(_3\) is added, it will react with the bromide ions (Br\(^-\)) to form silver bromide (AgBr), which precipitates.

**Calculation:**
- Moles of Br\(^-\) from \([Co(NH_3)_5SO_4]Br\) = 0.02 mol
- Therefore, moles of AgBr precipitated = 0.02 mol (1:1 ratio)

### Step 4: Remaining solution after AgNO\(_3\) addition
After the addition of AgNO\(_3\), the remaining solution contains:
- \([Co(NH_3)_5SO_4]Br\) (0.02 mol) without Br\(^-\)
- \([Co(NH_3)_5Br]SO_4\) (0.02 mol) with sulfate ions

### Step 5: Reaction with excess BaCl\(_2\)
When excess BaCl\(_2\) is added, it will react with the sulfate ions (SO\(_4^{2-}\)) to form barium sulfate (BaSO\(_4\)), which also precipitates.

**Calculation:**
- Moles of SO\(_4^{2-}\) from \([Co(NH_3)_5Br]SO_4\) = 0.02 mol
- Therefore, moles of BaSO\(_4\) precipitated = 0.02 mol (1:1 ratio)

### Final Result
The amounts of precipitated salts are:
- Precipitated AgBr: 0.02 mol
- Precipitated BaSO\(_4\): 0.02 mol

Thus, the final answer is that both precipitates are 0.02 mol.