Identify ionisable ions in each complex

The two compounds are:

$\left[\mathrm{Co}(\mathrm{NH_3})_5\mathrm{SO_4}\right]\mathrm{Br}$ and $\left[\mathrm{Co}(\mathrm{NH_3})_5\mathrm{Br}\right]\mathrm{SO_4}$

• In $\left[\mathrm{Co}(\mathrm{NH_3})_5\mathrm{SO_4}\right]\mathrm{Br}$, the $\mathrm{Br^-}$ is outside the coordination sphere, so it is ionisable.
• In $\left[\mathrm{Co}(\mathrm{NH_3})_5\mathrm{Br}\right]\mathrm{SO_4}$, the $\mathrm{SO_4^{2-}}$ is outside the coordination sphere, so it is ionisable.

Calculate moles of each compound taken

Each solution is $1\,\mathrm{L}$ and $0.02\,\mathrm{M}$.

So, moles of each complex salt: $n = MV = 0.02 \times 1 = 0.02\ \text{mol}$

Therefore:

• $0.02$ mol of $\left[\mathrm{Co}(\mathrm{NH_3})_5\mathrm{SO_4}\right]\mathrm{Br}$ gives $0.02$ mol $\mathrm{Br^-}$
• $0.02$ mol of $\left[\mathrm{Co}(\mathrm{NH_3})_5\mathrm{Br}\right]\mathrm{SO_4}$ gives $0.02$ mol $\mathrm{SO_4^{2-}}$

After mixing the two 1 L solutions

Total volume becomes: $1 + 1 = 2\,\mathrm{L}$

In this 2 L mixture, total free ions present are:

• $\mathrm{Br^-} = 0.02$ mol
• $\mathrm{SO_4^{2-}} = 0.02$ mol

The resulting solution is divided into two equal parts

So each part $X$ has volume: $1\,\mathrm{L}$

Hence each 1 L part contains half the total free ions:

• $\mathrm{Br^-} = \frac{0.02}{2} = 0.01\ \text{mol}$
• $\mathrm{SO_4^{2-}} = \frac{0.02}{2} = 0.01\ \text{mol}$

Reaction with excess $\mathrm{AgNO_3}$

$\mathrm{Ag^+ + Br^- \rightarrow AgBr(s)}$

Since $\mathrm{AgNO_3}$ is in excess, all free $\mathrm{Br^-}$ precipitates.

Therefore, moles of precipitate $Y = \mathrm{AgBr}$ are: $0.01\ \text{mol}$

Reaction with excess $\mathrm{BaCl_2}$

$\mathrm{Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4(s)}$

Since $\mathrm{BaCl_2}$ is in excess, all free $\mathrm{SO_4^{2-}}$ precipitates.

Therefore, moles of precipitate $Z = \mathrm{BaSO_4}$ are: $0.01\ \text{mol}$

Final answer

The number of moles of $Y$ and $Z$ respectively are: $0.01,\ 0.01$

So, the correct option is A.