# Preparation of Barium Hexaaluminate using co-precipitation reaction mechanism

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?

@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.

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.