Question

The amount of $H_{2}S$ , required to precipitate $1.69\,g.\, BaS$ from $BaCl_{2}$ solution is:

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

Answer 1

Answer: (C)

0.34 g

Answer 2

$\underset{(2 + 32 =34)}{BaCl _{2}+ H _{2} S} \rightarrow \underset{(137 + 32 =169)}{BaS} \downarrow+2 HCl$ $\because 169\, g BaS$ is obtained $34\, g \cdot H _{2} S$ $\therefore 1.69\,g\,BaS$ will be obtained by $=\frac{34 \times 1.69}{169}=0.34\, g\, H _{2} S$