`A 1.85 g` sample of mixture of `CuCl_(2)` and `CuBr_(2)` was dissolved in water and mixed thoroughly with `1.8 g ` portion of `AgCl`. After reaction, the solid which now dissoved contain `AgCl` and `AgBr` was filtered, dried and weighed to be `2.052 g.` What was the `%` by weight of `CuBr_(2)` in the mixture?

Answer

`34.18`

Explanation

Apply conservation of moles of silver before and after precipitate exchange reaction as :
`(1.8)/(143.5)=(x)/(188)+(2.052-x)/(143.5)`
where, `x` is mass of `AgBr` in mixed precipitate.
`implies x=1.064`
Also, moles of `CuBr_(2)=(1)/(2)` moles of `AgBr=(1)/(2)xx(x)/(188)`
implies Mass of `CuBr_(2)=(1)/(2)xx(x)/(188)xx223.5=0.6324`
Mass `%` of `CuBr_(2) = 34.18`