The minimum quantity in gram of H2S needed to precipitate 63.5 g of Cu2+ will be (Cu2+ + H2S → underset textBlackCu2S + H2)

Question

The minimum quantity in gram of H2S needed to precipitate 63.5 g of Cu2+ will be (Cu2+ + H2S → underset textBlackCu2S + H2) (A) 63.5 g (B) 31.75 g (C) 3 4 g (D) 2 g

Tardigrade Admin · Accepted Answer

underset63.5 gCu2+ + H2S → Cu2S +H2 1 mole of Cu2+ requires 1 mole of H2S or 63.5 g of Cu2+ requires 34 g of H2S [Molar mass of H2S = 2 + 32 = 34 g] So 1 mole H2S is required, i.e, 34 g

Q. The minimum quantity in gram of $H_{2} S$ needed to precipitate $63.5$ g of $C u^{2 +}$ will be
$(C u^{2 +} + H_{2} S \to Black C u_{2} S + H_{2})$

$63.5$ g

$31.75$ g

$34$ g

$2$ g

$63.5 g C u^{2 +} + H_{2} S \to C u_{2} S + H_{2}$
$1$ mole of $C u^{2 +}$ requires $1$ mole of $H_{2} S$
or $63.5 g$ of $C u^{2 +}$ 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, $34 g$

Q. The minimum quantity in gram of H2​S needed to precipitate 63.5 g of Cu2+ will be (Cu2++H2​S→BlackCu2​S​+H2​)

63.5 g

31.75 g

34 g

2 g