Hòa tan 9,61 gam hỗn hợp X gồm 3 kim loại Ba, Al và Fe vào nước

* Hướng dẫn giải

Đáp án C

$\begin{array}{l}+ Sơ đồ phản ứng:\\ \\ \underset{m_X=9,61 gam}{\underbrace{\left\{\begin{array}{l}Ba: x mol\\ Al\\ Fe: y mol\end{array}\right\}}} \stackrel{ }{\to } \left\{\begin{array}{l}{H}_2\uparrow : 0,12 mol\\ \underset{hỗn hợp Y}{\underbrace{\left\{\begin{array}{l}Al d¨o: y mol\\ Fe: z mol\end{array}\right\}}}Cu\downarrow : 0,11 mol\\ Ba(Al{O}_2{)}_2: x\end{array}\right. \\ \begin{array}{l}+ BTE cho \left(1\right): 2\underset{x}{\underbrace{n_{Ba}}}+3\underset{2x}{\underbrace{n_{Al p¨o}}}=2\underset{0,12}{\underbrace{n_{H_2}}}\Rightarrow x=0,03.\\ + \left\{\begin{array}{l}{m}_Y=m_X-m_{Ba}-m_{Al p¨o}\\ BTE cho \left(2\right): 3n_{Al}+2n_{Fe}=2n_{Cu}\end{array}\right.\Rightarrow \left\{\begin{array}{l}27y+56z=3,88\\ 3y+2z=0,11.2\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,04\\ y=0,05\end{array}\right.\\ \Rightarrow \%m_{Al}=\frac{27(0,06+0,04)}{9,61}=28,09\%\end{array}\\ \end{array}$