Question

The minimum quantity of $H_{2}S$ needed to precipitate $63.5g$ of $C{u}^{2+}$ will be nearly -

• A. $63.5g$
• B. $31.75g$
• C. $34.0g$
• D. $2.0g$

Answer 1

The correct option is C $34.0g$

Meq. of $H_{2}S=$ Meq. of $C{u}^{2+}$
$\therefore \frac{W_{H_{2}S}}{34.0/2}\times 1000=\frac{63.5}{63.5/2}\times 1000$
$\therefore W_{H_{2}S}=34.0g$
Hence, (c) is the correct answer.