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

Answer

To solve the problem, we need to determine the number of moles of precipitates Y (AgCl) and Z (BaSO4) that form when the given solutions are treated with excess silver nitrate and excess barium chloride, respectively.

### Step 1: Analyze the first complex compound
We have 0.02 moles of \([Co(NH_3)_5Br]Cl_2\). This compound dissociates in solution as follows:

\[ [Co(NH_3)_5Br]Cl_2 \rightarrow [Co(NH_3)_5Br]^{2+} + 2Cl^- \]

From this dissociation, for every mole of \([Co(NH_3)_5Br]Cl_2\), we produce 2 moles of \(Cl^-\). Therefore, the total moles of \(Cl^-\) produced from this compound is:

\[ \text{Moles of } Cl^- = 0.02 \text{ moles} \times 2 = 0.04 \text{ moles} \]

### Step 2: Analyze the second complex compound
Next, we have 0.02 moles of \([Co(NH_3)_5Cl]SO_4\). This compound dissociates as follows:

\[ [Co(NH_3)_5Cl]SO_4 \rightarrow [Co(NH_3)_5Cl]^{2+} + SO_4^{2-} \]

From this dissociation, for every mole of \([Co(NH_3)_5Cl]SO_4\), we produce 1 mole of \(SO_4^{2-}\). Therefore, the total moles of \(SO_4^{2-}\) produced from this compound is:

\[ \text{Moles of } SO_4^{2-} = 0.02 \text{ moles} \]

### Step 3: Precipitation with silver nitrate
When we treat the solution with excess silver nitrate (\(AgNO_3\)), the \(Ag^+\) ions will react with the \(Cl^-\) ions to form silver chloride (\(AgCl\)), which is a precipitate:

\[ Ag^+ + Cl^- \rightarrow AgCl \]

Since we have 0.04 moles of \(Cl^-\), the moles of \(AgCl\) formed will also be:

\[ \text{Moles of } AgCl = 0.04 \text{ moles} \]

Thus, \(Y = 0.04\) moles.

### Step 4: Precipitation with barium chloride
Next, when we treat the solution with excess barium chloride (\(BaCl_2\)), the \(Ba^{2+}\) ions will react with the \(SO_4^{2-}\) ions to form barium sulfate (\(BaSO_4\)), which is also a precipitate:

\[ Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4 \]

Since we have 0.02 moles of \(SO_4^{2-}\), the moles of \(BaSO_4\) formed will also be:

\[ \text{Moles of } BaSO_4 = 0.02 \text{ moles} \]

Thus, \(Z = 0.02\) moles.

### Final Answer
The number of moles of the precipitates Y and Z formed are:
- \(Y = 0.04\) moles (AgCl)
- \(Z = 0.02\) moles (BaSO4)

### Summary
The final answer is:
- \(Y = 0.04\)
- \(Z = 0.02\)