XCl2 (excess) + YCl2 XCl4 + Y ; YO 1/2 O2 + Y, Ore of Y would be : — Metallurgy and Isolation of Elements Chemistry Question
Question
XCl2 (excess) + YCl2 $\rightarrow$ XCl4 + Y $\downarrow$; YO $\xrightarrow{\Delta >400^\circ C}$ 1/2 O2 + Y, Ore of Y would be :
Answer: B
💡 Solution & Explanation
SnCl2 + HgCl2 $\rightarrow$ SnCl4 + Hg; (XCl2) (YCl2) (XCl4) (Y). HgO $\xrightarrow{\Delta >400^\circ C}$ Hg + 1/2 O2. HgS : Cinnabar
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