a) Gọi số mol Al, Fe là a, b (mol)
=> 27a + 56b = 11,1 (1)
\(n_{HCl}=\dfrac{60.36,5\%}{36,5}=0,6\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
a--->3a-------->a----->1,5a
Fe + 2HCl --> FeCl2 + H2
b--->2b------->b----->b
=> 3a + 2b = 0,6 (2)
(1)(2) => a = 0,1; b = 0,15
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{11,1}.100\%=24,32\%\\\%m_{Fe}=\dfrac{0,15.56}{11,1}.100\%=75,68\%\end{matrix}\right.\)
b) \(n_{H_2}=1,5a+b=\) 0,3 (mol)
=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
mdd sau pư = 11,1 + 60 - 0,3.2 = 70,5 (g)
\(\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,1.133,5}{70,5}.100\%=18,94\%\\C\%_{FeCl_2}=\dfrac{0,15.127}{70,5}.100\%=27,02\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
\(m_{HCl}=\dfrac{60\cdot36,5}{100}=21,9g\)
\(\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
x 3x 1,5x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y 2y y
\(\Rightarrow\left\{{}\begin{matrix}27x+56y=11,1\\3x+2y=0,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)
a)\(\%m_{Al}=\dfrac{0,1\cdot27}{11,1}\cdot100\%=24,32\%\)
\(\%m_{Fe}=100\%-24,32\%=75,68\%\)
b)\(\Sigma n_{H_2}=1,5x+y=1,5\cdot0,1+0,15=0,3mol\)
\(V_{H_2}=0,3\cdot22,4=6,72l\)
