Gọi $n_{Fe} = a(mol) ; n_{Al} = b(mol)$

$\Rightarrow 56a + 27b = 11(1)$

$Fe + 2HCl \to FeCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{H_2} = a + 1,5b = 8,96 : 22,4 = 0,4(2)$
Từ (1)(2) suy ra a = 0,1 ; b = 0,2

Vậy :

$\%m_{Fe} = \dfrac{0,1.56}{11}.100\% = 50,91\%$
$\%m_{Al} = 100\% -50,91\% = 49,09\%$

\(a,n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=11(1)\\ n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow 1,5x+y=0,4(2)\\ (1)(2)\Rightarrow x=0,2(mol);y=0,1(mol)\\ \Rightarrow \begin{cases} \%_{Al}=\dfrac{0,2.27}{11}.100\%=49,09\%\\ \%_{Fe}=100\%-49,09\%=50,91\% \end{cases}\\ b,\Sigma n_{HCl}=3x+2y=0,8(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,8}{2}=0,4(l)\)

\(n_{Al}=a\left(mol\right)\)

\(n_{Fe}=b\left(mol\right)\)

\(m=27a+56b=19.3\left(g\right)\left(1\right)\)

\(n_{H^+}=0.2\cdot2+0.2\cdot2.25\cdot2=1.3\left(mol\right)\)

\(2Al+6H^+\rightarrow2Al^{3+}+3H_2\)

\(Fe+2H^+\rightarrow Fe^{2+}+H_2\)

\(n_{H^+}=3a+2b=1.3\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.3,b=0.2\)

\(\%Al=\dfrac{0.3\cdot27}{19.3}\cdot100\%=41.96\%\)

\(\%Fe=58.04\%\)

\(b.\)

\(n_{H_2}=\dfrac{1}{2}n_{H^+}=0.65\left(mol\right)\)

Bảo toàn khối lượng : 

\(m_{Muối}=19.3+0.4\cdot36.5+0.45\cdot98-0.65\cdot2=76.7\left(g\right)\)

em cảm ơn ạ!

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

\(2A1+2NAOH+2H_2O-2NaA10_2+H_2O\)

\(AI_2O_3=2NaOH+2NaOHA10_2+H_2O\)

\(n_{AI}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,6=0,4\left(mol\right)\)

\(m_{AI}=27.0,4=10,8\left(gam\right);mAI_2O_3=31,2-10,8=20,4\left(gam\right)\)

Biết làm mỗi câu A

 Chỉ Có Al tác dụng với HCl sp tạo ra khí H₂

\(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
   0.4           \(\leftarrow\)                           0.6  (mol)

\(Al_2O_3+HCl\rightarrow AlCl_3+H_2O\)

n_Al=\(\dfrac{0.6\cdot2}{3}=0.4\left(mol\right)\)

m_Al=0.4*27=10.8(g)

=)) \(m_{Al_2O_3}=31.2-10.8=20.4\left(g\right)\)

\(\left\{{}\begin{matrix}\%Al=\dfrac{10.8}{31.2}\cdot100=34.62\%\\\%Al_2O_3=\dfrac{20.4}{31.2}\cdot100=65.38\%\end{matrix}\right.\)

\(n_{NaOH}=0,4.0,5=0,2\left(mol\right)\)

PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

Theo PT: \(n_{CH_3COOH}=n_{NaOH}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,2.60}{15,2}.100\%\approx78,95\%\\\%m_{C_2H_5OH}\approx21,05\%\end{matrix}\right.\)

\(2Al+6HCl\to 2AlCl_3+3H_2\\ Fe+2HCl\to FeCl_2+H_2\\ n_{HCl}=0,2.4=0,8(mol)\\ \Rightarrow \begin{cases} 56.n_{Fe}+27.n_{Al}=22\\ 2.n_{Fe}+3.n_{Al}=0,8 \end{cases}\Rightarrow \begin{cases} n_{Fe}=0,39(mol)\\ n_{Al}=0,007(mol) \end{cases}\\ \Rightarrow \begin{cases} \%m_{Fe}=\dfrac{0,39.56}{22}.100\%=99,27\%\\ \%m_{Al}=100\%-99,27\%=0,73\% \end{cases}\)

Đặt \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Fe}=y\end{matrix}\right.\) ( mol ) \(\rightarrow m_{hh}=27x+56y=5,54\left(g\right)\)  (1)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

 x                                                   1,5x    ( mol )

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

 y                                           y       ( mol )

\(n_{H_2}=1,5x+y=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)   (1)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,06\\y=0,07\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,06.27}{5,54}.100=29,24\%\\\%m_{Fe}=100-29,24=70,76\%\end{matrix}\right.\)

Đè thiếu rồi em ơi !