\(\begin{array}{l}
1)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
nNa = \dfrac{{2,3}}{{23}} = 0,1\,mol \Rightarrow n{H_2} = 0,05\,mol\\
\Rightarrow V{H_2} = 0,05 \times 22,4 = 1,12l\\
2)\\
nAlC{l_3} = \dfrac{{4,005}}{{133,5}} = 0,03\,mol\\
nNaOH = 0,1 \times 1 = 0,1\,mol\\
AlC{l_3} + 3NaOH \to 3NaCl + Al{(OH)_3}\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + {H_2}O\\
nAl{(OH)_3}(1) = nAlC{l_3} = 0,03\,mol\\
nAl{(OH)_3}(2) = 0,1 - 0,03 \times 3 = 0,01\,mol\\
mAl{(OH)_3} = (0,03 - 0,01) \times 78 = 1,56g\\
3)\\
2Al + 2NaOH + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}\\
nAl = \dfrac{{2,7}}{{27}} = 0,1\,mol \Rightarrow n{H_2} = 0,15\,mol\\
V{H_2} = \dfrac{{0,15 \times 22,4}}{2} = 1,68l
\end{array}\)