Gọi x = $n_{Fe2O3}$.
PT: Fe2O3 + 6HCl ---> 2FeCl3 + 3H2O
PƯ: x 2x (mol)
PT: Cu + 2FeCl3 ---> CuCl2 + 2FeCl2
PƯ: x 2x (mol)
Theo đề bài, ta có:
$m_{Fe2O3}$ + $m_{Cu}$ = $m_{hh}$
=> $m_{Fe2O3}$ + $m_{Cu PƯ}$ + $m_{Cu dư}$ = $m_{hh}$
=> 160x + 64x + 5,52 = 10
=> x = 0,02
% $m_{Fe2O3}$ = $\frac{160.0,02}{10}$ . 100 = 32%
% $m_{Cu}$ = $\frac{64.0,02+5,52}{10}$ . 100 = 68%