XCl2 (excess) + YCl2 → XCl4 + Y ↓; YO --Δ-->400° 1/2 O2 + Y; Ore of Y would be: — AITS & Test Series Chemistry Question

💡 Solution & Explanation

     2 2 4 2 2 4 SnCl HgCl SnCl Hg XCl YCl XCl Y    o 2 400 1 HgO Hg O 2     HgS : Cinnabar