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I am preparing Barium Hexaaluminate (BHA) using $\ce{Ba(NO3)2, Al(NO3)3}$, and $\ce{(NH4)2CO3}$. For $\pu{1 mol}$ of each $\ce{Ba(NO3)2}$ and $\ce{Al(NO3)4}$, I am using $\pu{1.5 mol}$ of $\ce{(NH4)2CO3}$.

Following are my questions:

  1. What temperature to select for reaction, according to some research papers we should use $\pu{60 ^\circ C}$. But the problem with keeping reaction temperature at $\pu{60 ^\circ C}$ will cause $\ce{(NH4)2CO3}$ to decompose into $\ce{NH3}$ and $\ce{CO2}$ and it will not take place in double displacement reaction.

  2. Can I replace $\ce{(NH4)2CO3}$ with some other carbonates?

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@Maurice, I always thought that barium nitrate and aluminium nitrate will react with ammonium carbonate and barium carbonate and aluminium hydroxide will precipitate down.

$$ \begin{align} \ce{2 Al(NO3)3 (aq) + 3 (NH4)2CO3 (aq) &-> Al2(CO3)3 (s) + 6 NH4NO3 (aq)}\tag{R1}\\ \ce{Ba(NO3)2 (aq) + (NH4)2CO3 (aq) &-> BaCO3 (s) + 2 NH4NO3 (aq)}\tag{R2} \end{align} $$

(As aluminium carbonate is not stable and readily decomposes to form aluminium hydroxide.)

After that, if we keep the ratio of barium and aluminium $1:12$ and raise the temperature above $\pu{1200 °C}$ by calcinating the dried precipitate, barium hexaaluminate forms.

Also, can you clarify what $\ce{BaAl6(OH)20}$ is, because according to my understanding the general formula of barium hexaluminate is $\ce{BaO.6(Al2O3)}.$

Reference

  1. Li, J. Q.; Wang, R. K.; Chen, C. Y. Preparation of Barium Aluminate with $\ce{BaCO3}$ and $\ce{Al(OH)3}$. AMR 2015, 1096, 156–160. DOI: 10.4028/www.scientific.net/AMR.1096.156.
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Heating a mixture of Barium and Aluminum salts with $\ce{(NH4)2CO3}$ is done on purpose, because $\ce{(NH4)2CO3}$ is decomposed at 58°C into $\ce{NH3}$ and $\ce{CO2}$ and $\ce{NH3}$ is the reagent wanted to react with Barium and Aluminium to make them precipitate together. This is due to the two successive reactions: $$\ce{NH3 + H2O <=> NH4+ + OH-}$$

$$\ce{20 OH^- + Ba^{2+} + 6Al^{3+} -> BaAl6(OH)20}$$

If you would have used $\ce{(NH4)2CO3}$ at ordinary temperature, it would have made a mixed precipitate of $\ce{BaCO3}$ and $\ce{Al(OH)3}$.

The oddity of this reaction is the fact that you use 1 mol Barium, 1 mole Aluminium, and 1.5 mole $\ce{(NH4)2CO3}$. Barium is in great excess, but it will not precipitate, as $\ce{Ba(OH)2}$ is quite soluble in water. I wonder why you use such a great excess of barium.

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