`n_{H_2}=(4,48)/(22,4)=0,2(mol)`; `n_{Fe}=(33,6)/56=0,6(mol)`

$PTHH: Fe+2HCl\longrightarrow FeCl_2+H_2$ `(1)`

$Fe_2O_3+3H_2\longrightarrow 2Fe+3H_2O$           `(2)`

Theo $PTHH$: `n_{Fe(1)}=n_{H_2(1)}=0,2(mol)→m_{Fe(1)}=0,2\times 56=11,2(g)`

`n_{Fe_2O_3(2)}=1/2 \times n_{Fe(2)}=,3(mol)→m_{Fe_2O_3(2)}=0,3\times 160=48(g)`

Do hh chia làm `2` phần bằng nhau nên `m_{Fe(1)}=m_{Fe(2)}; m_{Fe_2O3(1)}=m_{Fe_2O_3(2)}`

`→m_{tổng}=2\times (m_{Fe}+m_{Fe_2O_3})=118,4(g)`

`%m_{Fe}=(22,4)/(118,4)=18,9%`

`%m_{Fe_2O_3}=100%-18,9%=81,1%`

Đáp án:

\(\begin{array}{l}
\% {m_{Fe}} = \dfrac{{56 \times 0,4}}{{86,4}} \times 100\%  = 25,93\% \\
\% {m_{F{e_2}{O_3}}} = \dfrac{{160 \times 0,4}}{{86,4}} \times 100\%  = 74,07\% 
\end{array}\)

Giải thích các bước giải:

Phần 1:

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{{H_2}}} = 0,2mol\\
 \to {n_{Fe}} = {n_{{H_2}}} = 0,2mol\\
 \to {m_{Fe}} = 11,2g
\end{array}\)

Phần 2:

\(\begin{array}{l}
F{e_2}{O_3} + {H_2} \to 2Fe + 3{H_2}O\\
{n_{Fe}} = 0,6mol\\
 \to {n_{Fe}}(F{e_2}{O_3}) = 0,6 - 0,2 = 0,4mol\\
 \to {n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe}}(F{e_2}{O_3}) = 0,2mol
\end{array}\)

Suy ra hỗn hợp ban đầu ta có:
\(\begin{array}{l}
{n_{Fe}} = 0,4mol\\
{n_{F{e_2}{O_3}}} = 0,4mol\\
 \to {m_{Fe}} + {m_{F{e_2}{O_3}}} = 86,4g\\
 \to \% {m_{Fe}} = \dfrac{{56 \times 0,4}}{{86,4}} \times 100\%  = 25,93\% \\
 \to \% {m_{F{e_2}{O_3}}} = \dfrac{{160 \times 0,4}}{{86,4}} \times 100\%  = 74,07\% 
\end{array}\)