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The minimum quantity of H(2)S needed to ...

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

A

63.5gm

B

31.75gm

C

34gm

D

2.0gm

Text Solution

Verified by Experts

The correct Answer is:
C
Meq of `H_(2)S`=Meq of `Cu^(2+)`
therefore `(W_(H_(2)S))/(34//2)xx100=(64.5)/(63.5//2)xx100` therefore `W_(H_(2)S)=34 gm`
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