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 problem, we need to analyze the reactions that occur when 1 L of the mixture \(X\) is treated with excess \(AgNO_3\) and \(BaCl_2\).

### Step 1: Determine the composition of mixture \(X\)

The mixture \(X\) contains:
- \(0.02 \, \text{mol}\) of \([Co(NH_3)_5SO_4]Br\)
- \(0.02 \, \text{mol}\) of \([Co(NH_3)_5Br]SO_4\)

Since the total volume of the solution is \(2 \, L\), in \(1 \, L\) of the mixture \(X\), we have:
- \(0.01 \, \text{mol}\) of \([Co(NH_3)_5SO_4]Br\)
- \(0.01 \, \text{mol}\) of \([Co(NH_3)_5Br]SO_4\)

### Step 2: Reaction with \(AgNO_3\)

When \(1 \, L\) of mixture \(X\) is treated with excess \(AgNO_3\), the following reaction occurs:

\[ [Co(NH_3)_5SO_4]Br + AgNO_3 \rightarrow [Co(NH_3)_5SO_4]NO_3 + AgBr \]

In this reaction, only \([Co(NH_3)_5SO_4]Br\) reacts with \(AgNO_3\) to form \(AgBr\). Since we have \(0.01 \, \text{mol}\) of \([Co(NH_3)_5SO_4]Br\), it will produce \(0.01 \, \text{mol}\) of \(AgBr\).

Thus, the number of moles of \(Y\) (which is \(AgBr\)) produced is:
\[ \text{Moles of } Y = 0.01 \, \text{mol} \]

### Step 3: Reaction with \(BaCl_2\)

When \(1 \, L\) of mixture \(X\) is treated with excess \(BaCl_2\), the following reaction occurs:

\[ [Co(NH_3)_5Br]SO_4 + BaCl_2 \rightarrow [Co(NH_3)_5Br]Cl + BaSO_4 \]

In this case, only \([Co(NH_3)_5Br]SO_4\) reacts with \(BaCl_2\) to form \(BaSO_4\). Since we have \(0.01 \, \text{mol}\) of \([Co(NH_3)_5Br]SO_4\), it will produce \(0.01 \, \text{mol}\) of \(BaSO_4\).

Thus, the number of moles of \(Z\) (which is \(BaSO_4\)) produced is:
\[ \text{Moles of } Z = 0.01 \, \text{mol} \]

### Final Answer

The number of moles of \(Y\) and \(Z\) are:
- \(Y = 0.01 \, \text{mol}\)
- \(Z = 0.01 \, \text{mol}\)