Đáp án:

 12,8g và 33,6g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
CuO + {H_2} \to Cu + {H_2}O\\
b)\\
mhh = \dfrac{{16}}{{25\% }} = 64g\\
 \Rightarrow mF{e_2}{O_3} = 64 - 16 = 48g\\
nF{e_2}{O_3} = \dfrac{m}{M} = \dfrac{{48}}{{160}} = 0,3\,mol\\
nCuO = \dfrac{n}{M} = \dfrac{{16}}{{80}} = 0,2\,mol\\
nFe = 2nF{e_2}{O_3} = 0,6\,mol\\
 \Rightarrow mFe = n \times M = 0,6 \times 56 = 33,6g\\
nCu = nCuO = 0,2\,mol\\
mCu = n \times M = 0,2 \times 64 = 12,8g
\end{array}\)

$a/Fe_2O_3+H_2\to 2Fe+3H_2O$

$CuO+H_2\overset{t^o}\to Cu+H_2O$

$b/m_{CuO}=16.25\%=4g$

$⇒n_{CuO}=4/80=0,05mol$

$⇒n_{Cu}=n_{CuO}=0,05mol$

$⇒m_{Cu}=0,05.64=3,2g$

$m_{Fe_2O_3}=16-4=12g$

$⇒n_{Fe_2O_3}=32/160=0,2mol$

$⇒n_{Fe}=2.n_{Fe_2O_3}=2.0,2=0,4mol$

$⇒m_{Fe}=0,4.56=22,4g$