Bạn tham khảo!

a) Ta có 2 PTHH xảy ra:

$\text{1) Fe+2HCl$\rightarrow$FeCl2+H2}$

$\text{2) Fe2O3+6HCl$\rightarrow$2FeCl3+3H2O}$

b) 

Ta có $\text{4,48 lít khí H2}$ là $\text{V H2(1)}$

$\rightarrow$ $\text{nH2=$\dfrac{V}{22,4}$=$\dfrac{4,48}{22,4}$=0,2 mol}$

Dựa vào PTHH 1) $\text{nH2=nFe=0,2 mol}$

$\rightarrow$ $\text{mFe=n.M=0,2.56=11,2g}$

$\rightarrow$ Có $\text{mFe2O3=12,8-11,2=1,6g}$

Vậy $\text{%Fe=$\dfrac{11,2}{12,8}$.100=87,5%}$

$\text{%Fe2O3=100-87,5=12,5%}$

Em tham khảo nha  :

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
b)\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,2mol\\
{m_{Fe}} = 0,2 \times 56 = 11,2g\\
\% Fe = \dfrac{{11,2}}{{12,8}} \times 100\%  = 87,5\% \\
\% F{e_2}{O_3} = 100 - 87,5 = 12,5\% 
\end{array}\)