\(\begin{array}{l}
a)\\
2Al+6HCl\to 2AlCl_3+3H_2\\
Fe+2HCl\to FeCl_2+H_2\\
2Cu+O_2\xrightarrow{t^o}2CuO\\
4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\
3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\
b)\\
\rm Trong\,57,2g,X\,đặt\,n_{Cu}=x(mol);n_{Al}=y(mol);n_{Fe}=z(mol)\\
\rm Trong\,1,2mol,X\,đặt\,n_{Cu}=kx(mol);n_{Al}=ky(mol);n_{Fe}=kz(mol)\\
\to 64x+27y+56z=57,2(1)\\
n_{H_2}=\frac{26,88}{22,4}=1,2(mol)\\
n_{O_2}=\frac{89,6:5}{22,4}=0,8(mol)\\
Theo\,PT:\,1,5y+z=n_{H_2}=1,2(2)\\
Theo\,PT:\,\frac{1}{2}kx+\frac{3}{4}ky+\frac{2}{3}kz=n_{O_2}\\
\to k\left(\frac{1}{2}x+\frac{3}{4}y+\frac{2}{3}z\right)=0,8\\
Mat\,khac\,kx+ky+kz=k(x+y+z)=1,2\\
\to \dfrac{x+y+z}{\frac{1}{2}x+\frac{3}{4}y+\frac{2}{3}z}=\dfrac{1,2}{0,8}=1,5\\
\to x+y+z=0,75x+1,125y+z\\
\to 2x=y(3)\\
(1)(2)(3)\to \begin{cases} x=0,2\\ y=0,4\\ z=0,6 \end{cases}\\
\to \begin{cases} \%m_{Cu}≈22,38\%\\ \%m_{Al}≈18,88\%\\ \%m_{Fe}=58,74\% \end{cases}
\end{array}\)

Đáp án+ Giải thích các bước giải:

`a)`

`2Al+6HCl->2AlCl_3+3H_2`

`Fe+2HCl->FeCl_2+H_2`

`2Cu+O_2` $\xrightarrow{t^o}$ `2CuO`

`4Al+3O_2` $\xrightarrow{t^o}$ `2Al_2O_3`

`3Fe+4O_2` $\xrightarrow{t^o}$ `Fe_3O_4`

`b)`

`n_{H_2}={26,88}/{22,4}=1,2(mol)`

`V_{O_2}={89,6}/5=17,92(l)`

`n_{O_2}={17,92}/{22,4}=0,8(mol)`

`Gọi ` $\begin{cases} n_{Cu}:a^{(mol)}\\n_{Al}:b^{(mol)}\\n_{Fe}:c^{(mol)} \end{cases}$ `trong\ 57,2(g)\ X`

`=>64a+27b+56c=57,2\ (1)`

`Theo\ pt : 1,5b+c=1,2 (2)`

`Gọi ` $\begin{cases} n_{Cu}:ka^{(mol)}\\n_{Al}:kb^{(mol)}\\n_{Fe}:kc^{(mol)} \end{cases}$ `trong\ 1,2(mol)\ X`

`=>ka+kb+kc=1,2`

`<=>k(a+b+c)=1,2\ (2)`

`Theo\ pt : 1/{2}ka+3/{4}kb+2/{3}kc=0,8`

`<=>k(1/{2}a+3/{4}b+2/{3}c)=0,8`

`Ta\ co\ : {a+b+c}/{1/{2}a+3/{4}b+2/{3}c}={1,2}/{0,8}`

`=>2a-b=0\ (3)`

`(1)\ , (2)\ ,(3)=>` $\begin{cases} a=0,2\\b=0,4\\c=0,6 \end{cases}$

`%m_{Cu}={0,2.64.100%}/{57,2}~~22,4%`

`%m_{Al}={0,4.27.100%}/{57,2}~~18,9%`

`%m_{Fe}=100%-22,4%-18,9%=58,7%`