Đáp án:

$a,\\ \%m_{Zn}=54,62\%\\ \%m_{Al}=45,38\%\\ a=13,92\\ b,\ Fe_3O_4$

Giải thích các bước giải:

$a,\\ n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ Đặt\ \begin{cases}n_{Zn}=x(mol)\\n_{Al}=y(mol)\end{cases}\\ \Rightarrow 65x+27y=11,9\ (1)\\ Zn+2HCl\to ZnCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \xrightarrow{\text{Từ PTHH}}\ n_{H_2}=n_{Zn}+\dfrac{3}{2}.n_{Al}=x+\dfrac{3}{2}y(mol)\\ \Rightarrow x+\dfrac{3}{2}y=0,4\ (2)\\ \xrightarrow{\text{Từ (1),(2)}}\begin{cases}x=0,1\\y=0,2\end{cases}\\ \%m_{Zn}=\dfrac{0,1.65}{11,9}.100\%=54,62\%\\ \%m_{Al}=\dfrac{0,2.27}{11,9}.100\%=45,38\%\\ \xrightarrow{\text{BTNT H}}\ n_{H_2O}=n_{H_2}=0,4(mol)\\ \Rightarrow m_{H_2O}=0,4.18=7,2(g)\\ \xrightarrow{\text{BTKL}}\ m_{H_2}+m_Y=m_{Fe}+m_{H_2O}\\ \Rightarrow 0,4.2+11,52+a=19,04+7,2\\ \Rightarrow a=13,92\\ b,\\ n_{FeO}=\dfrac{11,52}{72}=0,16(mol)\\ n_{Fe}=\dfrac{19,04}{56}=0,34(mol)\\ \xrightarrow{\text{BTNT Fe}}\ n_{FeO}+x.n_{Fe_xO_y}=n_{Fe}\\ \Rightarrow 0,16+x.n_{Fe_xO_y}=0,34\\ \Rightarrow n_{Fe_xO_y}=\dfrac{0,18}{x}(mol)\\ \Rightarrow M_{Fe_xO_y}=13,92:\dfrac{0,18}{x}=\dfrac{232}{3}x(g/mol)\\ \Rightarrow 56x+16y=\dfrac{232}{3}x\\ \Rightarrow \dfrac{64}{3}x=16y\\ \Rightarrow \dfrac{x}{y}=\dfrac{3}{4}\\ \to CTHH:\ Fe_3O_4$