Đáp án+Giải thích các bước giải:
$\rm \%Fe=70\%\rightarrow \%O=100\%-70\%=30\%\\CTDC:Fe_xO_y\\\dfrac{\%Fe}{\%O}=\dfrac{70\%}{30\%}=\dfrac{7}{3}\\\rightarrow \dfrac{56x}{16y}=\dfrac{7}{3}\\\rightarrow \dfrac{x}{y}=\dfrac{2}{3}\\\rightarrow CTHH:Fe_2O_3\\b)\text{Các phản ứng xảy ra}:\\CuO+H_2\xrightarrow {t^o}Cu+H_2O(1)\\Fe_xO_y+yH_2\xrightarrow{t^o} xFe+yH_2O(2)\\Fe+2HCl\rightarrow FeCl_2+H_2↑(3)\\n_{H_2(3)}=\dfrac{0,448}{22,4}=0,02(mol)\\n_{Fe}=n_{H_2(3)}=0,02(mol)\\\rightarrow m_{Fe}=0,02.56=1,12g\\\rightarrow m_{Cu}=1,76-1,12=0,64g\\n_{Cu}=\dfrac{0,64}{64}=0,1(mol)\\\xrightarrow{\text{Theo PTHH(1)}}n_{CuO}=n_{Cu}=0,1(mol)\\\rightarrow m_{CuO}=0,1.80=8g\\\rightarrow m_{O(Oxit)}=2,4-8=1,6g\\\dfrac{m_{Fe}}{m_O}=\dfrac{1,12}{0,48}\\\rightarrow \dfrac{56x}{16y}=\dfrac{1,12}{0,48}\\\rightarrow \dfrac{x}{y}=\dfrac{2}{3}\\Oxit:Fe_2O_3$