`1L,0.02M` solution of `[Co(NH_(3))_(5)SO_(4)]` Br is mixed with `1L,0.02M` solution of `[Co(NH_(3))_(5)]SO_(4)`.The resulting solution is divided into two equal parts `(X)` and treated with excess of `AgNO_(4)` solution and `BaCl_(2)`
`1` L Solution `(X)+AgNO_(3)` solution (excess `)rarr Y` `1L` Solution `(X)+AgNO_(3)` solution (excess) `rarr Y` `1` L Solution `(X)+BaCl_(2)` solution (excess) `rarr Y` The number of moles of `Y` and `Z` respectively are

Options

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

Answer

To solve the problem, we need to analyze the reactions occurring in the solutions and how they interact with the reagents added (AgNO3 and BaCl2).

### Step-by-Step Solution:

1. **Identify the Compounds:**
- The first solution is `[Co(NH3)5SO4]Br`, which contains cobalt in a +3 oxidation state and has a bromide ion as a counter ion.
- The second solution is `[Co(NH3)5]SO4`, which contains cobalt in a +3 oxidation state and has a sulfate ion as a counter ion.

2. **Calculate the Total Moles of Each Compound:**
- For both solutions, since they are 1L and 0.02M, the moles of each compound are:
- Moles of `[Co(NH3)5SO4]Br` = 1L × 0.02 mol/L = 0.02 moles
- Moles of `[Co(NH3)5]SO4` = 1L × 0.02 mol/L = 0.02 moles

3. **Mixing the Solutions:**
- When the two solutions are mixed, the total volume becomes 2L, and the total moles of each compound remain the same:
- Total moles of `[Co(NH3)5SO4]Br` = 0.02 moles
- Total moles of `[Co(NH3)5]SO4` = 0.02 moles

4. **Dividing the Solution:**
- The resulting solution is divided into two equal parts (1L each). Thus, each part (X) contains:
- Moles of `[Co(NH3)5SO4]Br` in X = 0.01 moles
- Moles of `[Co(NH3)5]SO4` in X = 0.01 moles

5. **Reaction with Excess AgNO3:**
- When excess AgNO3 is added to solution X, it will react with the bromide ion from `[Co(NH3)5SO4]Br` to form AgBr (a precipitate):
- Moles of AgBr formed = 0.01 moles (from the 0.01 moles of bromide)
- The cobalt complex `[Co(NH3)5]SO4` does not react with AgNO3.

6. **Calculating Moles in Solution Y:**
- After the reaction with AgNO3, solution Y contains:
- Cobalt complexes: 0.01 moles of `[Co(NH3)5SO4]` and 0.01 moles of `[Co(NH3)5SO4]Br` (but the Br is now precipitated as AgBr)
- Moles of Y = 0.01 (from `[Co(NH3)5]SO4`) + 0 (from `[Co(NH3)5SO4]Br` since it's precipitated) = 0.01 moles

7. **Reaction with Excess BaCl2:**
- When excess BaCl2 is added to the other half of the solution X, it will react with the sulfate ion from `[Co(NH3)5]SO4` to form BaSO4 (a precipitate):
- Moles of BaSO4 formed = 0.01 moles (from the 0.01 moles of sulfate)
- The cobalt complex `[Co(NH3)5SO4]Br` does not react with BaCl2.

8. **Calculating Moles in Solution Z:**
- After the reaction with BaCl2, solution Z contains:
- Cobalt complexes: 0.01 moles of `[Co(NH3)5SO4]Br` (but the SO4 is now precipitated as BaSO4)
- Moles of Z = 0 (from `[Co(NH3)5]SO4` since it's precipitated) + 0.01 (from `[Co(NH3)5SO4]Br`) = 0.01 moles

### Final Answer:
- The number of moles of Y = 0.01 moles
- The number of moles of Z = 0.01 moles