Hòa tan hoàn toàn m gam hỗn hợp X gồm Al và Al2O3 trong 200,0 ml dung dịch NaOH 2M

* Hướng dẫn giải

Đáp án B

$\begin{array}{l}+ Sơ đồ phản ứng:\\ \underset{m=(2,7+102x) gam}{\underbrace{\left\{\begin{array}{l}Al: \frac{2n_{H_2}}{3}=0,1 mol\\ A{l}_2{O}_3: x mol\end{array}\right\}}}\stackrel{NaOH: 0,4 mol}{\to }\underset{dd Y}{\underbrace{\left\{\begin{array}{l}{n}_{NaAl{O}_2}=0,1+2x\\ n_{NaOH}=0,3-2x\end{array}\right\}}}+\underset{0,15 mol}{\underbrace{H_2\uparrow }}\\ \underset{dd Y}{\underbrace{\left\{\begin{array}{l}NaAl{O}_2\\ NaOH\end{array}\right\}}}+\left[\begin{array}{l}0,3y mol HCl \left(TN1\right)\\ 0,7y mol HCl \left(TN2\right)\end{array}\right. \stackrel{ }{\to } \underset{\frac{2,7+102x}{78} mol}{\underbrace{m gam Al{\left(OH\right)}_3\downarrow }}+...\\ \Rightarrow \left\{\begin{array}{l}TN1: Al(OH{)}_3 chưa bị \tan \\ TN2: Al(OH{)}_3 đã bị \tan \end{array}\right.\Rightarrow \left\{\begin{array}{l}TN1: n_{H^{+}}=n_{O{H}^{-}}+n_{Al{\left(OH\right)}_3}\\ TN2: n_{H^{+}}=n_{O{H}^{-}}+n_{Al{O_2}^{-}}+3(n_{Al{O_2}^{-}}-n_{Al{\left(OH\right)}_3})\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}0,3y=(0,3-2x)+\frac{2,7+102x}{78}\\ 0,7y=(0,3-2x)+(0,1+2x)+3\left((0,1+2x)-\frac{2,7+102x}{78}\right)\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,05\\ y=0,1\end{array}\right.\\ \Rightarrow m=7,8 gần nhất với 8\end{array}$