# Question #21c0a

Dec 2, 2015

$\text{BaO"_2 * 8"H"_2"O}$

#### Explanation:

You can only get hydrogen peroxide, ${\text{H"_2"O}}_{2}$, by reacting ice-cold diluted hydrochloric acid with barium peroxide octahydrate.

The reaction will produce barium chloride, ${\text{BaCl}}_{2}$, water, and hydrogen peroxide

${\text{BaO"_2 * 8"H"_2"O"_text((s]) + 2"HCl"_text((aq]) -> "BaCl"_text(2(aq]) + 8"H"_2"O"_text((l]) + "H"_2"O}}_{\textrm{2 \left(a q\right]}}$

The water comes from the water of crystallization stored in the octahydrate. The problem with using hydrochloric acid to get hydrogen peroxide is that the resulting barium salt is soluble in aqueous solution.

The barium cation impurities will actually accelerate the hydrogen peroxide's decomposition, which is something you want to avoid when preparing hydrogen peroxide solutions.

None of the other oxides will react with hydrochloric acid to produce hydrogen peroxide. For example, manganese dioxide will react with hydrochloric acid to produce chlorine gas

${\text{MnO"_text(2(aq]) + 4"HCl"_text((aq]) -> "MnCl"_text(2(aq]) + "Cl"_text(2(g]) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

Here chlorine is being oxidized to chlorine gas and manganese is being reduced from a $+ 4$ oxidation state to a $+ 2$ oxidation state.