Đáp án:

$\%m_{Zn}=70,65\%\\ \%m_{Al}=29,35\%\\ a=0,4$

Giải thích các bước giải:

$a,\\ n_{H_2\ \rm{TN1}}=\dfrac{8,96}{22,4}=0,4(mol)\\ n_{H_2\ \rm{TN2}}=\dfrac{11,2}{22,4}=0,5(mol)\\ \rm{Ta\ có} :\ \begin{cases}m_{X\ \rm{TN1}}=m_{X\ \rm{TN2}}\\n_{HCl\ \rm{TN1}}=n_{HCl\ \rm{TN2}}\\n_{H_2\ \rm{TN1}}<n_{H_2\ \rm{TN2}}\end{cases}\\ \to \rm{TN1:\ X\ dư,\ HCl\ hết}$

Nếu $a(mol)\ HCl\to 0,4(mol)\ H_2$

Thì $2a(mol)\ HCl\to 0,8(mol)\ H_2$

$\rm{Mà}\ 0,8>0,5\\ \to \rm{TN2:\ HCl\ dư,\ X\ hết}$

$b,\\ TN2:\\ Đặt\ \begin{cases}n_{Zn}=x(mol)\\n_{Al}=y(mol)\end{cases}\\ \Rightarrow 65x+27y=18,4\ (1)\\ Zn+2HCl\to ZnCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \xrightarrow{\text{Từ PTHH}}\ n_{H_2}=n_{Zn}+\dfrac{3}{2}.n_{Al}=x+\dfrac{3}{2}y(mol)\\ \Rightarrow x+\dfrac{3}{2}y=0,5\ (2)\\ \xrightarrow{\text{Từ (1),(2)}}\begin{cases}x=0,2\\y=0,2\end{cases}\\ \Rightarrow \%m_{Zn}=\dfrac{0,2.65}{18,4}.100\%=70,65\%\\ \Rightarrow \%m_{Al}=\dfrac{0,2.27}{18,4}.100\%=29,35\%\\ TN1:\\ n_{HCl}=2.n_{H_2}=0,8(mol)\\ \Rightarrow 2a=0,8\\ \Rightarrow a=0,4$