Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Cu}=c\left(mol\right)\end{matrix}\right.\left(a,b,c>0\right)\)
=> 27a + 56b + 64c = 3,31 (1)
\(n_{H_2}=\dfrac{0,784}{22,4}=0,035\left(mol\right)\)
PTHH: `2Al + 6HCl -> 2AlCl_3 + 3H_2`
a--------------------------->1,5a
`Fe + 2HCl -> FeCl_2 + H_2`
b-------------------------->b
=> 1,5a + b = 0,035 (2)
Trong 0,12 mol X có: \(\left\{{}\begin{matrix}n_{Al}=ak\left(mol\right)\\n_{Fe}=bk\left(mol\right)\\n_{Cu}=ck\left(mol\right)\end{matrix}\right.\left(k>0\right)\)
=> ak + bk + ck = 0,12 (3)
PTHH: `2Al + 3Cl_2 -t^o-> 2AlCl_3`
ak------------------->ak
`2Fe + 3Cl_2 -t^o-> 2FeCl_3`
bk---------------------->bk
`Cu + Cl_2 -t^o-> CuCl_2`
ck------------------->ck
=> 133,5ak + 162,5bk + 135ck = 17,27 (4)
(1),(2),(3),(4) => \(\left\{{}\begin{matrix}a=0,01\\b=0,02\\c=0,03\\k=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{3,31}.100\%=8,16\%\\\%m_{Fe}=\dfrac{0,02.56}{3,31}.100\%=33,84\%\\\%m_{Cu}=\left(100-8,16-33,84\right)\%=58\%\end{matrix}\right.\)