\(\begin{array}{l}
a)\\
F{e_x}{O_y} + y{H_2} \to xFe + y{H_2}O\\
CuO + {H_2} \to Cu + {H_2}O\\
n{H_2} = \dfrac{{14,112}}{{22,4}} = 0,63\,mol \Rightarrow n{H_2}O = n{H_2} = 0,63\,mol\\
BTKL:\\
m + m{H_2} = mhh + m{H_2}O \Rightarrow m = mhh + m{H_2}O - m{H_2}\\
\Rightarrow m = 32,16 + 0,63 \times 18 - 0,63 \times 2 = 42,24g\\
b)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
n{H_2} = \dfrac{{8,064}}{{22,4}} = 0,36\,mol \Rightarrow nFe = n{H_2} = 0,36\,mol\\
mCu = 32,16 - 0,36 \times 56 = 12g\\
\Rightarrow nCu = \dfrac{{12}}{{64}} = 0,1875\,mol\\
nCuO = nCu = 0,1875\,mol \Rightarrow mCuO = 15g\\
mF{e_x}{O_y} = 42,24 - 15 = 27,24g\\
mO = 27,24 - 0,36 \times 56 = 7,08g\\
\Rightarrow nO = \dfrac{{7,08}}{{16}} = 0,4425\,mol\\
nFe:nO = 0,36:0,4425 \approx 3:4 \Rightarrow CT:F{e_3}{O_4}
\end{array}\)
