Chemistry - Salt Analysis Question with Solution | TestHub

Because Gas (B) is colourless and turns acidified ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ solution green, it appears to be ${\mathrm{H}}_2\mathrm{S}$.

${\mathrm{H}}_2\mathrm{S}$ gas is obtained by the reaction of dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$ on A , thus A must be a sulphide.

ZnS sulphide is white thus it indicates that A is ZnS .

$\underset{(\mathrm{A})}{\mathrm{ZnS}}+{\mathrm{H}}_2{\mathrm{SO}}_4$ (dil.) $\to \underset{(\mathrm{C})}{{\mathrm{ZnSO}}_4}+\underset{(\mathrm{B})}{{\mathrm{H}}_2\mathrm{S}}\uparrow$ $3\underset{\text{ (B) }}{{\mathrm{H}}_2\mathrm{S}}+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\to {\mathrm{K}}_2{\mathrm{SO}}_4+\underset{\text{ (Green) }}{{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3}+7{\mathrm{H}}_2\mathrm{O}+\underset{\text{ (D) }}{3\mathrm{S}}$

Green colored compound is ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$.