# Question #862f2

Feb 4, 2016

Not by the reaction between potassium chloride, potassium dichromate, and concentrated sulfuric acid.

#### Explanation:

Chlorine gas, ${\text{Cl}}_{2}$, is not obtained by heating potassium chloride, $\text{KCl}$, potassium dichromate, ${\text{K"_2"Cr"_2"O}}_{7}$, and concentrated sulfuric acid, ${\text{H"_2"SO}}_{4}$.

This reaction is actually called the chromyl chloride test for the chloride anion, ${\text{Cl}}^{-}$.

The idea is that when you heat a compound that contains chloride (a solid salt, not a solution that contains the solvated anion) with potassium dichromate and concentrated sulfuric acid, the reaction produces chromyl chloride, ${\text{CrO"_2"Cl}}_{2}$, a red fuming liquid.

The balanced chemical equation for this reaction is

$4 {\text{KCl"_text((s]) + "K"_2"Cr"_2"O"_text(7(s]) + "H"_2"SO"_text(4(aq]) stackrel(color(red)(Delta)color(white)(aa))(->) 6"KHSO"_text(4(aq]) + 2"CrO"_2"Cl"_text(2(g]) + 3"H"_2"O}}_{\textrm{\left(l\right]}}$

Mind you, this is kind of an overall description of the reaction.

Here's what's actually going on - the sulfuric acid reacts with the potassium chloride and potassium dichromate separately to produce hydrochloric acid, $\text{HCl}$, and chromium trioxide, ${\text{CrO}}_{3}$, respectively.

$\text{KCl" + "H"_2"SO"_4 -> "KHSO"_4 + "HCl}$

$\text{K"_2"Cr"_2"O"_7 + 2"H"_2"SO"_4 -> 2"KHSO"_4 + 2"CrO"_3 + "H"_2"O}$

It is then the reaction between hydrochloric acid and chromium trioxide that produces chromyl chloride.

$2 \text{HCl" + "CrO"_3 -> "CrO"_2"Cl"_2 + "H"_2"O}$

If you add these equations together (and balance them out), you'll get the chemical equation that describes the overall reaction.

I think that some chlorine gas can actually be produced here as a side-product of the reaction, but even if that happens this reaction will still qualify as your answer.

Here's a very cool video detailing the reaction

All the other reactions produce chlorine gas, ${\text{Cl}}_{2}$, as a major product.

Potassium permanganate, ${\text{KMnO}}_{4}$, will react with concentrated hydrochloric acid to form potassium chloride, $\text{KCl}$, mangane(II) chloride, ${\text{MnCl}}_{2}$, chlorine gas, and water.

http://socratic.org/questions/how-to-balance-an-equation-in-its-molecular-form-eg-kmno4-hcl-gives-kcl-mncl2-h2

Manganese dioxide, ${\text{MnO}}_{2}$, will react with concentrated hydrochloric acid to produce manganese(II) chloride, chlorine gas, and water

Finally, potassium chloride will react with fluorine gas to form chlorine gas and potassium fluoride, $\text{KF}$
${\text{KCl"_text((s]) + "F"_text(2(g]) -> 2"KF"_text((s]) + "Cl}}_{\textrm{2 \left(g\right]}}$