$a,n_X=\dfrac{10,08}{22,4}=0,45(mol)\\ Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O\\ Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\\ CuO+H_2\xrightarrow{t^o}Cu+H_2O\\ CuO+CO\xrightarrow{t^o}Cu+CO_2\\ \Rightarrow n_{O(oxit)}=n_X=0,45(mol)\\ \Rightarrow m=m_{\text{kim loại}}+m_{O(oxit)}=24,8+0,45.16=32(g)\\ b,M_Z=\dfrac{28.1+28.1}{2}=28(g/mol)\\ \dfrac{M_X}{M_Z}=0,35 \Rightarrow M_X=0,35.28=9,8(g/mol)\\ \text{Đặt }n_{H_2}=a(mol);n_{CO}=b(mol)\\ \Rightarrow M_X=\dfrac{2a+28b}{a+b}=9,8\\ \Rightarrow 7,8a=18,2b\\ \Rightarrow \dfrac{a}{b}=\dfrac{7}{3}\\ \Rightarrow \dfrac{n_{H_2}}{n_{CO}}=\dfrac{7}{3}\\ \Rightarrow \dfrac{V_{H_2}}{V_{CO}}=\dfrac{7.22,4}{3.22,4}=\dfrac{7}{3}$