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If solid $\ce{Hg2Cl2}$ is placed in concentrated $\ce{HCl}$, will it react?
If so are the products $\ce{Hg2^2+}$, $\ce{H2}$, and $\ce{Cl2}$ ?

$$\ce{Hg2Cl2(s) + 2HCl(aq) -> Hg2^2+ + 2Cl2(g) + H2(g)}$$

I am teaching predicting products on a Chemistry I level, and I am not familiar with the subtleties of mercury chemistry. The short answer is "no reaction" because our "rules" state that a double displacement must occur with both reactants in aqueous solution. But I am seeking the reality behind the theory.

The entropy change would favor the reaction since it yields 4 mole of gasses from 1 mole of solid and 2 mole of solution.

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    $\begingroup$ I think you didn't thought-out well this reaction... $\endgroup$ – Mithoron Feb 10 '16 at 17:04
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    $\begingroup$ Your equation is not charge-balanced. Please edit your post to indicate what reaction you're interested in. $\endgroup$ – Todd Minehardt Feb 10 '16 at 17:22
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    $\begingroup$ I think the reaction would be: $$\ce{Hg2Cl2(s) + 2HCl(aq) -> HgCl4^{2-} + 2H^+ + Hg(s)}$$ Heating would definitely give this reaction. $\endgroup$ – MaxW Feb 11 '16 at 5:01

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