Question

$\text{Black}\left(A\right)+H_{2}S{O}_4\to \text{gas}\left(B\right)+\left(C\right)$
$\left(B\right)+(C{H}_{3}COO{)}_{2}Pb\to \text{black ppt}(D)$
$\left(C\right)+K_3\left[Fe\right(CN{)}_6]\to \text{blue}(E)$
$\left(C\right)$ also decolorises acidified $KMn{O}_4$.

Identify $\left(A\right)$.

• A. $F{e}_{2}S$
• B. $PbS$
• C. $FeS{O}_4$
• D. $FeS$

Answer 1

The correct option is D $FeS$

$\text{Black}\left(A\right)+H_{2}S{O}_4\to \text{gas}\left(B\right)+\left(C\right)$
$FeS+H_{2}SO4\to FeS{O}_4+H_{2}S$. So the gas may be $H_2$ and C would be $FeS{O}_4$ and A is FeS.
$\left(B\right)+(C{H}_{3}COO{)}_{2}Pb\to \text{black ppt}(D)$
$Pb(CH3COO{)}_2+H_{2}S\to PbS+2C{H}_{3}COOH$. So the balck ppt. D would be Pbs.
$\left(C\right)+K_3\left[Fe\right(CN{)}_6]\to \text{blue}(E)$
$FeS{O}_4+K_3\left[\right(Fe\left(CN{)}_6\right)]\to KFe(Fe\left(CN{)}_6\right)+K_{2}S{O}_4$.So $4KFe\left(Fe\right(CN{)}_6)$ is blue colour E.
$\left(C\right)$ also decolorises acidified $KMn{O}_4$. Identify $\left(A\right)$.
$FeS{O}_4$ declorises acidified $KMn{O}_4$.
So, $A$ would be $FeS$. Hence, option $D$ is correct.