A mixture `x` containing 0.02 mol of `[Co(NH_(3))_(5) SO_(4)]Br` and 0.02 mol of `[Co(NH_(3))_(5)Br]SO_(4)` was prepared in `2 L` of solution.
`1 L` of mixture `X + ` excess `AgNO_(3) rarr Y`
`1 L` of mixture `X +` excess `BaCl_(2) rarr Z`
The number of moles of `Y` and `Z` are

Options

A. 0.01,0.01
B. 0.02,0.01
C. 0.01,0.02
D. 0.02,0.02

Explanation

To solve the question step by step, we will analyze the given information and apply the relevant concepts from coordination chemistry and stoichiometry.

### Step 1: Understand the Mixture
We have a mixture \( X \) containing:
- 0.02 mol of \([Co(NH_3)_5 SO_4]Br\) (let's call this Complex 1)
- 0.02 mol of \([Co(NH_3)_5Br]SO_4\) (let's call this Complex 2)

The total volume of the solution is 2 L.

### Step 2: Calculate Moles in 1 L of Mixture
Since the total amount of the mixture is 2 L, when we take 1 L of this mixture, the moles of each complex will be halved:
- Moles of Complex 1 in 1 L = \( \frac{0.02}{2} = 0.01 \) mol
- Moles of Complex 2 in 1 L = \( \frac{0.02}{2} = 0.01 \) mol

### Step 3: Reaction with Excess \( AgNO_3 \)
When we add excess \( AgNO_3 \) to the 1 L of mixture:
- \( AgNO_3 \) will react with Complex 1, which has the bromide ion (Br) as the counter ion.
- The reaction will produce \( AgBr \) as a precipitate.

The stoichiometry of the reaction is:
\[ [Co(NH_3)_5 SO_4]Br + AgNO_3 \rightarrow [Co(NH_3)_5 SO_4]NO_3 + AgBr \]

Since we have 0.01 mol of Complex 1, it will react completely to form 0.01 mol of \( AgBr \).

### Step 4: Calculate Moles of \( Y \)
Thus, the number of moles of \( Y \) (which is \( AgBr \)) formed is:
- Moles of \( Y = 0.01 \) mol

### Step 5: Reaction with Excess \( BaCl_2 \)
Next, we take another 1 L of the mixture and add excess \( BaCl_2 \):
- \( BaCl_2 \) will react with Complex 2, which has the sulfate ion (SO₄) as the counter ion.
- The reaction will produce \( BaSO_4 \) as a precipitate.

The reaction is:
\[ [Co(NH_3)_5Br]SO_4 + BaCl_2 \rightarrow [Co(NH_3)_5Br]Cl_2 + BaSO_4 \]

Again, since we have 0.01 mol of Complex 2, it will react completely to form 0.01 mol of \( BaSO_4 \).

### Step 6: Calculate Moles of \( Z \)
Thus, the number of moles of \( Z \) (which is \( BaSO_4 \)) formed is:
- Moles of \( Z = 0.01 \) mol

### Final Answer
The number of moles of \( Y \) and \( Z \) are both \( 0.01 \) mol.

### Summary
- Moles of \( Y = 0.01 \) mol
- Moles of \( Z = 0.01 \) mol