Hỗn hợp X gồm a mol Al và b mol Na. Hỗn hợp Y

* Hướng dẫn giải

$$\begin{gathered} * TN1: \\ \left\{\begin{array}{l}Al: a mol\\ Na: b mol\end{array} \stackrel{H2O}{\to }\left\{\begin{array}{l}NaAl{O}_2: bmol\\ H_2 \\ Al\end{array}\right.\right. \\ BT e: n_{Na} + 3. n_{Pb} = \\ b + 3b = 2.n_{H2} = 0,48 \\ \to b = 0,12 \\ * TN2: \\ \left\{\begin{array}{l}Al: 0,12 mol\\ Na: a mol\end{array} \stackrel{H2O}{\to } \left\{\begin{array}{l}NaAl{O}_2: 0,12 mol\\ NaOH: 0,03 mol\end{array}\right.+ ...\right. \\ a=n_{Na}=n_{NaOH}+n_{NaAlO2}=0,15 \\ \to n_{Al(TN1,TN2)} = 0,27 mol \\ \to m_{Al} = 7,29 g \end{gathered}$$