- Thí nghiệm 1: $\underbrace{\overbrace{Al}^{a\,mol},\,\overbrace{Na}^{b\,mol}}_{\text{hn h }\!\!\hat{\mathrm{i}}\!\!\text{ p X}}\xrightarrow{+\ {{H}_{2}}O}{{X}_{1}}+\underbrace{{{H}_{2}}}_{0,24\ mol}+$ \[\underbrace{A{{l}_{d}}}_{m\ gam}\]. Ta có: ${{n}_{Na}}=b=\frac{{{n}_{{{H}_{2}}}}}{2}=0,12\ mol$

- Thí nghiệm 2:$\underbrace{\overbrace{Al}^{0,12\,mol},\,\overbrace{Na}^{a\,mol}}_{dung\text{ dch Y}}\xrightarrow{+\ {{H}_{2}}O}\,\underbrace{\overbrace{NaOH}^{0,03\,mol},\,\overbrace{NaAl{{O}_{2}}}^{0,12\,mol}}_{dung\text{ dch}\,{{\text{Y}}_{2}}}$. Khi đó: $\xrightarrow{BT:\ Na}a=0,15\ mol$

$\Rightarrow {{m}_{Al}}=27.(a+b)=7,29\ gam$