Chemistry - Salt Analysis Question with Solution | TestHub

Step 1: Pb²⁺ reacts with dilute HCl to form A, which is a white precipitate of Lead(II) chloride (PbCl₂).

Step 2: Passing gas C through PbCl₂ produces precipitate D (PbS). In qualitative analysis, hydrogen sulfide (H₂S) is the gas used to precipitate lead as a sulfide.

Step 3: In the presence of excess KCl, lead forms the $[PbC{l}_4]$²⁻ complex. Reacting this with H₂S forms the intermediate precipitate E (Pb₂Cl₂S), which yields PbS and PbCl₂ upon dilution.

Conclusion: Gas C is H₂S.

The correct option is (B) H₂S.