a, \(n_{HCl}=0,1.1,5=0,15\left(mol\right);n_{H_2}=\dfrac{1,288}{22,4}=0,0575\left(mol\right)\)
Ta có: \(\dfrac{0,15}{8}>\dfrac{0,0575}{4}\) ⇒ HCl dư
⇒ tính số mol theo H2
Fe + 2HCl ----> FeCl2 + H2
x x x x
2Al + 6HCl ----> 2AlCl3 + 3H2
y 3y 2y 1,5y
Ta có: hệ pt: \(\left\{{}\begin{matrix}56x+27y=1,795\\x+1,5y=0,0575\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,025\left(mol\right)\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{0,02.56.100\%}{1,795}=62,4\%;\%m_{Al}=100\%-62,4\%=37,6\%\)
b, Trong dd X có HCl dư, FeCl2 và AlCl3
\(n_{HCldư}=0,15-0,02-3.0,025=0,055\left(mol\right)\)
HCl + KOH ----> KCl + H2O
0,055 0,055
2KOH + FeCl2 ----> Fe(OH)2 + 2KCl
0,04 0,02
3KOH + AlCl3 -----> Al(OH)3 + 3KCl
0,15 0,05
\(n_{KOH}=0,055+0,04+0,15=0,245\left(mol\right)\)
\(\Rightarrow V_{KOH}=\dfrac{0,245}{1}=0,245\left(l\right)=245\left(ml\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
Fe + 2HCl ---> FeCl2 + H2
a a a
2Al + 6HCl ---> 2AlCl3 + 3H2
b b 1,5b
\(n_{H_2}=\dfrac{1,228}{22,4}=0,0575\left(mol\right)\)
Hệ pt \(\left\{{}\begin{matrix}56a+27b=1,795\\a+1,5b=0,0575\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,02\left(mol\right)\\b=0,025\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,02.56=1,12\left(g\right)\\\%m_{Al}=0,025.27=0,675\left(g\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{1,12}{1,795}=62,4\%\\\%m_{Al}=100\%-62,4\%=37,6\%\end{matrix}\right.\)
PTHH:
FeCl2 + 2KOH ---> Fe(OH)2 + 2KCl
0,02 0,04 0,02
AlCl3 + 3KOH ---> Al(OH)3 + 3KCl
0,025 0,075 0,025
\(\rightarrow V_{ddKOH}=\dfrac{0,075+0,04}{1}=0,115\left(l\right)\)
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