Hòa tan m gam hỗn hợp X gồm Al và Na có tỉ lệ mol 1:2 vào nước dư, thu được

* Hướng dẫn giải

Đáp án B

HH $\left\{\begin{array}{l}N\text{a}:2x\\ Al:x\end{array}\right.$

$$\begin{gathered} 2Na+2H_{2}O\to 2NaOH+H_2 \\ 2x2x x \\ 2\text{A}l+2H_{2}O+2NaOH\to 2NaAl{O}_2+3H_2 \\ x 2\text{x\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1,5x \end{gathered}$$

Sau phản ứng, dung dịch chứa NaOH và NaAlO₂.

$$\begin{gathered} n_{H_2}=2,5\text{x}=0,2\to x=0,08. \\ \to m=m_{Al}+m_{Na}=0,08.27+0,16.23=5,84\left(g\right). \end{gathered}$$