Cinnabar $(H g S)$ is heated in air, it is first converted into __ which is then reduced to _ on further heating.

H g O

H g

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H g O

H g_{2} O

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H g_{2} O

H g

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H g_{2} O

H g O

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Solution

The correct option is B $H g O$ , $H g$
Cinnabar $(H g S)$ is heated in air, it is first converted into $(H g O)$ which is then reduced to Hg on further heating.
$2 H g S + 3 O_{2} Δ - \to 2 H g O + 2 S O_{2}$
$2 H g O Δ - \to 2 H g + O_{2}$

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Similar questions

Given that,

$Z n + \frac{1}{2} O_{2} \to Z n O + 35.25 K J$

$H g O \to H g + \frac{1}{2} O_{2} - 9.11 K J$

The heat of the reaction $Z n + H g O \to Z n O + H g$ is

Z n + 1 / 2 O_{2} \to Z n O + 35.25 k J . H g O \to H g + 1 / 2 O_{2} + 9.11 k J .

Z n + H g O \to Z n O + H g

H g O (s) H e a t - -- \to H g (l) + O_{2} (g)

H g O (s) + H_{2} (g) \to H g (l) + H_{2} O (l)

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