Đáp án đúng: B
8,1 (g).
$\begin{array}{l}\text{2Na}\,\text{+}\,\text{2}{{\text{H}}_{\text{2}}}\text{O}\,\xrightarrow{{}}\,\text{2NaOH}\,\,\text{+}\,\,{{\text{H}}_{\text{2}}}\\\,\,\text{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\xrightarrow{{}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{x}\,\xrightarrow{{}}\,\,\frac{\text{x}}{\text{2}}\\\text{Al}\,\text{+}\,\text{NaOH}\,\text{+}{{\text{H}}_{\text{2}}}\text{O}\,\xrightarrow{{}}\,\text{NaAl}{{\text{O}}_{\text{2}}}\,\text{+}\,\frac{\text{3}}{\text{2}}{{\text{H}}_{\text{2}}}\\\text{x}\,\,\,\leftarrow \,\,\,\text{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\xrightarrow{{}}\,\,\,\,\,\,\,\,\,\,\,\frac{\text{3x}}{\text{2}}\\\Rightarrow \,\frac{x}{2}\,+\,\frac{3x}{2}\,=\,0,3\,mol\,\Rightarrow \,x=\,0,15\,mol\\\Rightarrow \,{{n}_{Na}}\,=\,0,15;\,{{n}_{Al}}\,=\,0,45\end{array}$
