Mercury and oxygen react to form mercury(II) oxide, like this: 2 Hg(l)+02(g)--HgO(s) At a certain temperature, a chemist finds that a 6.9 L reaction vessel containing a mixture of mercury, oxygen, and mercury(II) oxide at equilibrium has the following composition: compound amount Hg 16.9 g O 10.9 g HgO 23.8 g Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Kc = 20

Explanation:

We have the equilibrium:

2 Hg (l) + O₂ ( g) ⇄  HgO (s)

Kc = 1/ [O₂]

The key here is to remember that  pure solids and liquids do not enter into the calculation for the equilibrium expression, and Hg is a pure liquid and HgO is a solid

So what we need to do to solve this question is to calculate the concentration of oxygen at equilibrium. We are given its mass, and the volume so we are equipped to calculate the concentration of oxygen as follows:

[O₂] = # moles O₂ / V

# moles O₂ = mass / molar mass = 10.9 g / 32g/mol = 0.34 mol

[O₂] = 0.34 mol / 6.9 L = 0.049 M

⇒ Kc = 1 / 0.049 = 20 ( rounded to 2 significant figures )

yes

explanation:

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1.43 x 10^-11 j the guy commented the awnser just making sure u see it .