Question

The minimum quantity in gm of H₂S needed to precipitate 63.5 gm of Cu²⁺ ions will be nearly

(a) 63.5 gm

(b) 31.75 gm

(c) 34 gm

(d) 20 gm

Answer 1

Correct option is (c) 34 gm

Cu²⁺+ H₂S → CuS + 2H⁺

The quantity of H₂S required = \(\frac{34 \times 63.5}{63.5}\) = 34 g