a) Vì chất rắn sau pư + dd NaOH -> H2 => chất rắn sau phản ứng gồm Al dư, Fe, Al2O3
PTHH: \(8Al+3Fe_3O_4\xrightarrow[]{t^o}4Al_2O_3+9Fe\) (1)
Phần 1: \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\) (2)
\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\) (3)
\(Fe+2HCl\rightarrow FeCl_2+H_2\) (4)
Phần 2: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\) (5)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (6)
\(Fe+2HCl\rightarrow FeCl_2+H_2\) (7)
b) `-` Xét phần 1: \(\left\{{}\begin{matrix}n_{H_2\left(3\right)}=\dfrac{1,176}{22,4}=0,0525\left(mol\right)\\n_{H_2\left(4\right)}=\dfrac{1,008}{22,4}=0,045\left(mol\right)\end{matrix}\right.\)
Theo PT (3): \(n_{Al}=\dfrac{2}{3}n_{H_2\left(3\right)}=0,035\left(mol\right)\)
Theo PT (4): \(n_{Fe}=n_{H_2\left(4\right)}=0,045\left(mol\right)\)
`-` Xét phần 2:
Đặt hệ số tỉ lệ: \(\dfrac{P_2}{P_1}=k\left(k>0\right)\)
`=>` \(\left\{{}\begin{matrix}n_{Al}=0,035k\left(mol\right)\\n_{Fe}=0,045k\left(mol\right)\end{matrix}\right.\)
Theo PT (6), (7): \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{6,552}{22,4}=0,2925\left(mol\right)\)
`=> 0,0525k + 0,045k = 0,2925`
`=> k = 3`
`-` Xét hh ban đầu: \(\dfrac{hhbđ}{P_1}=\dfrac{P_1+P_2}{P_1}=\dfrac{P_1+3P_1}{P_1}=4\)
`=>` \(\left\{{}\begin{matrix}n_{Al}=0,035.4=0,14\left(mol\right)\\n_{Fe}=0,045.4=0,18\left(mol\right)\end{matrix}\right.\)
Theo PT (1): \(n_{Al\left(pư\right)}=\dfrac{8}{9}n_{Fe}=0,16\left(mol\right)\)
`=>` \(m_A=\left(0,14+0,16\right).27+0,18.56=18,18\left(g\right)\)
`=>` \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,3.27}{18,18}.100\%=44,55\%\\\%m_{Fe}=100\%-44,55\%=55,45\%\end{matrix}\right.\)
