6) Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) => 56a + 27b = 11 (1)
\(m_{ddHCl}=500.1,1=550\left(g\right)\Rightarrow n_{HCl}=\dfrac{550.5,31\%}{36,5}=0,8\left(mol\right)\)
PTHH:
Fe + 2HCl ---> FeCl2 + H2
a--->2a--------->a------->a
2Al + 6HCl ---> 2AlCl3 + 3H2
b--->3b--------->b------->1,5b
=> 2a + 3b = 0,8 (2)
(1), (2) => a = 0,1; b = 0,2
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{0,1.56+0,2.27}.100\%=50,9\%\\\%m_{Al}=100\%-50,9\%=49,1\%\end{matrix}\right.\)
b) V = 22,4.(0,1 + 0,2.1,5) = 8,96 (l)
mdd sau phản ứng = 550 + 11 - 2.(0,1 + 0,2.1,5) = 560,2 (g)
=> \(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,1.127}{560,2}.100\%=2,27\%\\C\%_{AlCl_3}=\dfrac{0,2.133,5}{560,2}.100\%=4,77\%\end{matrix}\right.\)
