Question

The minimum quantity of $H_{2}S$ needed to precipitate $64.5g$ of $C{u}^{2+}$ will be nearly.

• A. $63.5g$
• B. $31.75g$
• C. $34g$
• D. $2.0g$

Answer 1

Answer: (C)

Meq. of $H_{2}S=$ Meq. of $C{u}^{2+}$
$\therefore \frac{{}^w{H}_{2}S}{342}\times 1000=\frac{63.5}{63.52}\times 1000$
$\therefore {}^w{H}_{2}S=34g$