`0.02` mole of `[Co(NH_(3))_(5)Br]Cl_(2)` and 0.02 mole of `[Co(NH_(3))_(5) Cl] SO_(4)` are present in 200 cc of a solution X . The number of moles of the precipitates Y and Z that are formed when the solution X is treated with excess silver nitrate and excess barium chloride are respectively

Options

A. 0.02 , 0.02
B. 0.01 , 0.02
C. 0.02 , 0.04
D. 0.04 , 0.02

Explanation

To solve the problem, we need to determine the number of moles of precipitates formed when the solution X, containing two coordination compounds, is treated with excess silver nitrate and excess barium chloride.

### Step-by-Step Solution:

1. **Identify the Coordination Compounds:**
- We have 0.02 moles of \([Co(NH_3)_5Br]Cl_2\).
- We also have 0.02 moles of \([Co(NH_3)_5Cl]SO_4\).

2. **Determine the Chloride Ions from Each Compound:**
- The compound \([Co(NH_3)_5Br]Cl_2\) has 2 chloride ions per formula unit. Therefore, from 0.02 moles:
\[ \text{Chloride ions from } [Co(NH_3)_5Br]Cl_2 = 0.02 \, \text{moles} \times 2 = 0.04 \, \text{moles} \]
- The compound \([Co(NH_3)_5Cl]SO_4\) has 1 chloride ion per formula unit. Therefore, from 0.02 moles:
\[ \text{Chloride ions from } [Co(NH_3)_5Cl]SO_4 = 0.02 \, \text{moles} \times 1 = 0.02 \, \text{moles} \]

3. **Total Chloride Ions in Solution:**
- Adding the chloride ions from both compounds:
\[ \text{Total chloride ions} = 0.04 \, \text{moles} + 0.02 \, \text{moles} = 0.06 \, \text{moles} \]

4. **Precipitation with Silver Nitrate:**
- When excess silver nitrate (\(AgNO_3\)) is added, it reacts with chloride ions to form silver chloride (\(AgCl\)):
\[ Ag^+ + Cl^- \rightarrow AgCl \]
- Since we have 0.06 moles of chloride ions, the number of moles of \(AgCl\) precipitate formed will also be:
\[ \text{Moles of } AgCl = 0.06 \, \text{moles} \]

5. **Precipitation with Barium Chloride:**
- Next, we consider the sulfate ions from the compound \([Co(NH_3)_5Cl]SO_4\). This compound contributes 1 sulfate ion per formula unit:
\[ \text{Sulfate ions from } [Co(NH_3)_5Cl]SO_4 = 0.02 \, \text{moles} \]
- When excess barium chloride (\(BaCl_2\)) is added, it reacts with sulfate ions to form barium sulfate (\(BaSO_4\)):
\[ Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4 \]
- Therefore, the number of moles of \(BaSO_4\) precipitate formed will be:
\[ \text{Moles of } BaSO_4 = 0.02 \, \text{moles} \]

6. **Final Result:**
- The number of moles of precipitate \(Y\) (from \(AgCl\)) is 0.06 moles.
- The number of moles of precipitate \(Z\) (from \(BaSO_4\)) is 0.02 moles.

### Conclusion:
The number of moles of the precipitates \(Y\) and \(Z\) formed are 0.06 moles and 0.02 moles respectively.