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?


2 Answers 2


@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 $\ce{BaO.6(Al2O3)}.$


  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.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.