Gọi số mol $CuO$ và $Fe_2O_3$ là $x,y$ (mol)
$n_{H_2}=\dfrac{13,44}{22,4}=0,6(mol)$
$m_{hh}=80x+160y=40\\↔2x+4y=1\quad (1)$
PTHH:
$Cu+H_2→Cu+H_2O\\\,\,\,x\quad \quad x\quad\quad\quad\quad\quad\quad\quad\quad(mol)$
$Fe_2O_3+H_2→2Fe+3H_2O\\\quad y\quad \quad \quad 3y\quad\quad\quad\quad\quad\quad\quad (mol) $
$\sum n_{H_2}=x+3y=0,6(mol)\quad (2)$
Từ (1)(2) $→$ Hpt $\begin{cases}2x+4y=1\\x+3y=0,6\end{cases}↔\begin{cases}x=0,3(mol)\\y=0,1(mol)\end{cases}$
$↔\begin{cases}n_{CuO}=0,3(mol)\\n_{Fe_2O_3}=0,1(mol)\end{cases}\\→\begin{cases}m_{CuO}=0,3.80=24(g)\\m_{Fe_2O_3}=16(g)\end{cases}$
Vậy $m_{CuO}=24g,\,m_{Fe_2O_3}=16g$
b/ $\% m_{CuO}=\dfrac{24}{40}=0,6=60\%\\\% m_{Fe_2O_3}=\dfrac{16}{40}=0,4=40\%$
Vậy $\%m_{CuO}=60\%,\,\%m_{Fe_2O_3}=40\%$
