# Question #ac1a8

Dec 29, 2016

This is not really a practical reaction...........

#### Explanation:

$A g C l \left(s\right) + K B r \left(s\right) \rightarrow A g B r \left(s\right) + K C l \left(a q\right)$

$\text{Moles of silver chloride}$ $=$ $\frac{12.0 \cdot g}{143.32 \cdot g \cdot m o {l}^{-} 1} = 0.0837 \cdot m o l .$

$\text{Moles of potassium bromide}$ $=$ $\frac{13.0 \cdot g}{119.0 \cdot g \cdot m o {l}^{-} 1} = 0.109 \cdot m o l .$

Clearly, there is sufficient bromide anion to effect metathesis, and should the reaction go to completion, then $0.0837 \cdot m o l$ $K B r$ would eventually precipitate, which constitutes a mass of $0.0837 \cdot m o l \times 187.77 \cdot g \cdot m o {l}^{-} 1 = 15.72 \cdot g$. I think you can calculate the mass of the excess potassium bromide.

Had this reaction been performed, you would see the white precipitate of $A g C l$ change to the cream-coloured $A g B r$. While silver halides are both quite insoluble, $A g B r$ is more insoluble than $A g C l$, and this would drive the reaction to the right as written. The problem with this reaction is that both silver salts are photo-active, and would reduce to give metallic silver as a dark precipitate.