Question

The minimum quantity in gram of $H_{2}S$ needed to precipitate $63.5$ g of $Cu^{2+}$ will be
$(Cu^{2+} + H_{2}S \to \underset{\text{Black}}{Cu_{2}S} + H_{2})$

• A. $63.5$ g
• B. $31.75$ g
• C. $3 4$ g
• D. $2$ g

Answer 1

Answer: (C)

$\underset{63.5\,g}{Cu^{2+}} + H_{2}S \to Cu_{2}S +H_{2}$
$1$ mole of $Cu^{2+}$ requires $1$ mole of $H_{2}S$
or $63.5\, g$ of $Cu^{2+}$ requires $34\, g$ of $H_{2}S$
[Molar mass of $H_{2}S = 2 + 32 = 34\, g]$
So $1$ mole $H_{2}S$ is required, i.e, $34\, g$