Đáp án:

 a=1,02

Giải thích các bước giải:

\(\begin{array}{l}
PTHH:\\
2Na + 2{H_2}O \to 2NaOH + {H_2}(1)\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl(2)\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O(3)\\
2Al{(OH)_3} \xrightarrow{t^0} A{l_2}{O_3} + 3{H_2}O(4)\\
{n_{Na}} = \dfrac{{2,3}}{{23}} = 0,1\,mol\\
{n_{AlC{l_3}}} = 0,1 \times 0,3 = 0,03\,mol\\
{n_{NaOH(1)}} = {n_{Na}} = 0,1\,mol\\
\dfrac{{{n_{NaOH}}}}{3} > {n_{AlC{l_3}}} \Rightarrow NaOH \text{ dư }\\
{n_{NaOH(2)}} = 3{n_{AlC{l_3}}} = 0,09\,mol\\
{n_{Al{{(OH)}_3}(2)}} = {n_{AlC{l_3}}} = 0,03\,mol\\
{n_{NaOH(3)}} = {n_{Al{{(OH)}_3}(3)}} = 0,1 - 0,09 = 0,01\,mol\\
{n_{Al{{(OH)}_3}(4)}} = 0,03 - 0,01 = 0,02\,mol\\
{n_{A{l_2}{O_3}}} = \dfrac{{0,02}}{2} = 0,01\,mol\\
{m_{A{l_2}{O_3}}} = 0,01 \times 102 = 1,02g
\end{array}\)