Question

Consider a mixture $X$ which is made by dissolving $0.4$ mol of $[\mathrm{Co(NH_3)_5SO_4}]Br$ and $0.4$ mol of $[\mathrm{Co(NH_3)_5Br}]SO_4$ in water to make $4$ L of solution. When $2$ L of mixture $X$ is allowed to react with excess $\mathrm{AgNO_3}$, it forms precipitate $Y$. The rest $2$ L of mixture $X$ reacts with excess $\mathrm{BaCl_2}$ to form precipitate $Z$. Which of the following statements is CORRECT?

• A. $Y$ is $\mathrm{BaSO_4}$ and $Z$ is $\mathrm{AgBr}$
• B. $0.1$ mol of $Y$ is formed
• C. $0.2$ mol of $Z$ is formed
• D. $0.4$ mol of $Z$ is formed

Hint:

Only counter ions take part in precipitation reactions; ligands inside coordination sphere do not.

Answer 1

The Correct Option is C

Step 1: Understanding the complexes. 
$[\mathrm{Co(NH_3)_5SO_4}]Br$ gives Br$^-$ as counter ion in solution. 
$[\mathrm{Co(NH_3)_5Br}]SO_4$ gives SO$_4^{2-$} as counter ion in solution. 
Step 2: Calculating moles in 2 L solution. 
Total volume = $4$ L 
So, $2$ L contains half the moles. 
\[ \text{Moles of each salt in 2 L} = \frac{0.4}{2} = 0.2 \] Step 3: Reaction with AgNO$_3$. 
Only free Br$^-$ reacts with $\mathrm{AgNO_3}$ to form $\mathrm{AgBr}$. 
Moles of $\mathrm{AgBr}$ formed: 
\[ 0.2\,\text{mol} \] Step 4: Reaction with BaCl$_2$. 
Only free SO$_4^{2-}$ reacts with $\mathrm{BaCl_2}$ to form $\mathrm{BaSO_4}$. 
Moles of $\mathrm{BaSO_4}$ formed: 
\[ 0.2\,\text{mol} \] Step 5: Final conclusion. 
The correct statement is that $0.2$ mol of $Z$ is formed.