X C l 2 e x c e s s + Y C l 2 → X C l 4 + Y ↓ YO → > 400 o C Δ 1 2 O 2 +Y , Ore of Y would be

The reaction sequence is as follows S n C l 2 ( X C l 2 ) + H g C l 2 ( Y C l 2 ) → S n C l 4 ( X C l 4 ) + H g ( Y ) HgO → > 400 o C Δ Hg + 1 2 O 2 HgS → Cinnabar

Chemistry - General Principles and Processes of Isolation of Metals Question with Solution | TestHub

ChemistryGeneral Principles and Processes of Isolation of MetalsGeneralEasy2 minPYQ_2020

ChemistryEasysingle choice

$X C l_{2} (e x c e s s) + Y C l_{2} \to X C l_{4} + Y ↓$
$YO \to_{> 400^{o} C}^{Δ} \frac{1}{2} O_{2} +Y$ , Ore of Y would be

Options:

Answer:

Solution:

The reaction sequence is as follows
$\underset{(X C l_{2})}{S n C l_{2}} + \underset{(Y C l_{2})}{H g C l_{2}} \to \underset{(X C l_{4})}{S n C l_{4}} + \underset{(Y)}{H g}$
$HgO \to_{> 400^{o} C}^{Δ} Hg + \frac{1}{2} O_{2}$
$HgS \to Cinnabar$

Stream:NTA_ABHYASSubject:ChemistryTopic:General Principles and Processes of Isolation of MetalsSubtopic:General

⏱ 2mℹ️ Source: PYQ_2020

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