Question

The amountof $H_{2}S$ required to precipitate $1.69$g $BaS$ from $BaC{l}_2$ solution is :

• A. $3.4g$
• B. $0.24g$
• C. $0.34g$
• D. $0.17g$

Answer 1

The correct option is C $0.34g$

$BaC{l}_2+H_{2}S⟶BaS+2HCl$
$1.69$ $g$
For $1$ $mole$ of $H_{2}S$ produce $1$ $mole$ of $BaS$
So, $1$ $mole$ $H_{2}S⟶1$ $mole$ $BaS$
$34$ $g$ $169$ $g$
$\left(x\right)⟵1.69$ $g$
$⟹x=\frac{1.69\times 34}{169}=0.34$ $g$
Amount of $H_{2}S$ needed $=0.34$ $g$