Question

The minimum quantity in grams of $H_{2}S$ needed to precipitate $63.5\text{g}$ of $C{u}^{2+}$ ion will be:
(molar mass of $Cu=63.5\text{g/mol}$)

$C{u}^{2+}+H_{2}S\to CuS+2H^{+}$

• A. $63.5\text{g}$
• B. $41.7\text{g}$
• C. $34.0\text{g}$
• D. $22.4\text{g}$

Answer 1

The correct option is C $34.0\text{g}$

$C{u}^{2+}+H_{2}S\to CuS+H_2\left(g\right)$

$63.5\text{g}$ of $C{u}^{2+}$ = 1 mol of $C{u}^{2+}$
1 mol of $H_{2}S$ is required to precipitate 1 mol of $C{u}^{2+}$
So, 1 mol $H_{2}S$ = $1\text{mol}\times 34\text{g/mol}=34\text{g}$.