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I know that dysprosium oxide is insoluble in water.

I was told that I need to add concentrated $\ce{HCl}$ to the compound and then evaporate the $\ce{HCl}$ off. From looking at the reaction equation, I can see that I will get dysprosium chloride and water as products.

Once I'm able to dissolve the $\ce{Dy2O3}$, how is it that I'm supposed to evaporate off the excess $\ce{HCl}$? Do I just let it sit, or do I need to gently heat the solution? How can I tell when all the excess $\ce{HCl}$ is gone and I can start adding water to obtain the concentration of dysprosium chloride that I need?

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DyCl3 is often prepared by the "ammonium chloride route," starting from either Dy2O3 or hydrated chloride or oxychloride.[2][3] or DyCl3·6H2O.[4] These methods produce (NH4)2[DyCl5]:

    10 NH4Cl + Dy2O3 → 2 (NH4)2[DyCl5] + 6 NH3 + 3 H2O

    DyCl3·6H2O + 2 NH4Cl → (NH4)2[DyCl5] + 6 H2O

The pentachloride decomposes thermally according to the following equation:

    (NH4)2[DyCl5] → 2 NH4Cl + DyCl3

The thermolysis reaction proceeds via the intermediacy of (NH4)[Dy2Cl7].

Treating Dy2O3 with aqueous HCl produces hydrated chloride (DyCl3·6H2O). This salt cannot be rendered anhydrous by heating. Instead one obtains an oxychloride.

source: wikipedia

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