Question #862f2

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Question #862f2

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Answer 1 · 2016-02-04T22:45:59

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

Explanation:

Chlorine gas, ${Cl}_{2}$ $"Cl"_2$ , is not obtained by heating potassium chloride, $KCl$ $"KCl"$ , potassium dichromate, $K_{2} {Cr}_{2} O_{7}$ $"K"_2"Cr"_2"O"_7$ , and concentrated sulfuric acid, $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ .

This reaction is actually called the chromyl chloride test for the chloride anion, ${Cl}^{-}$ $"Cl"^(-)$ .

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, ${CrO}_{2} {Cl}_{2}$ $"CrO"_2"Cl"_2$ , a red fuming liquid.

The balanced chemical equation for this reaction is

$4 {KCl}_{(s]} + K_{2} {Cr}_{2} O_{7(s]} + H_{2} {SO}_{4(aq]} Δ a a - - \to 6 {KHSO}_{4(aq]} + 2 {CrO}_{2} {Cl}_{2(g]} + 3 H_{2} O_{(l]}$ $4"KCl"_text((s]) + "K"_2"Cr"_2"O"_text(7(s]) + "H"_2"SO"_text(4(aq]) stackrel(color(red)(Delta)color(white)(aa))(->) 6"KHSO"_text(4(aq]) + 2"CrO"_2"Cl"_text(2(g]) + 3"H"_2"O"_text((l])$

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, $HCl$ $"HCl"$ , and chromium trioxide, ${CrO}_{3}$ $"CrO"_3$ , respectively.

$KCl + H_{2} {SO}_{4} \to {KHSO}_{4} + HCl$ $"KCl" + "H"_2"SO"_4 -> "KHSO"_4 + "HCl"$

$K_{2} {Cr}_{2} O_{7} + 2 H_{2} {SO}_{4} \to 2 {KHSO}_{4} + 2 {CrO}_{3} + H_{2} O$ $"K"_2"Cr"_2"O"_7 + 2"H"_2"SO"_4 -> 2"KHSO"_4 + 2"CrO"_3 + "H"_2"O"$

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

$2 HCl + {CrO}_{3} \to {CrO}_{2} {Cl}_{2} + H_{2} O$ $2"HCl" + "CrO"_3 -> "CrO"_2"Cl"_2 + "H"_2"O"$

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, ${Cl}_{2}$ $"Cl"_2$ , as a major product.

Potassium permanganate, ${KMnO}_{4}$ $"KMnO"_4$ , will react with concentrated hydrochloric acid to form potassium chloride, $KCl$ $"KCl"$ , mangane(II) chloride, ${MnCl}_{2}$ $"MnCl"_2$ , 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, ${MnO}_{2}$ $"MnO"_2$ , will react with concentrated hydrochloric acid to produce manganese(II) chloride, chlorine gas, and water

http://socratic.org/questions/hcl-is-added-to-following-oxides-which-one-would-give-h2o2-1-mno2-2-pbo2-3-bao2-

Finally, potassium chloride will react with fluorine gas to form chlorine gas and potassium fluoride, $KF$ $"KF"$

${KCl}_{(s]} + F_{2(g]} \to 2 {KF}_{(s]} + {Cl}_{2(g]}$ $"KCl"_text((s]) + "F"_text(2(g]) -> 2"KF"_text((s]) + "Cl"_text(2(g])$

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Question #862f2

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