The minimum quantity of `H_(2)`S needed to precipitate 64.5 g of `Cu^(2+)` will be nearly.

Options

A. (a)63.5 g
B. (b)31.75 g
C. (c )34 g
D. (d)2.0 g

Answer

c

Explanation

Meq. of `H_(2)S= "Meq. Of " Cu^(2+)`
`:. W_(H_(2)S)/(34//2)xx1000=63.5/(63.5//2)xx1000 :. W_(H_(2)S=34 g`