a) PTHH : \(3H_2+Fe_2O_3-t^o->2Fe+3H_2O\)
Ta có : \(m_{CR\left(giảm\right)}=m_{O\left(lay.di\right)}\)
=> \(m_{O\left(lay.di\right)}=32-24,8=7,2\left(g\right)\)
=> \(n_{O\left(lay.di\right)}=\frac{7,2}{16}=0,45\left(mol\right)\)
Theo pthh : \(n_{H_2\left(pứ\right)}=n_{O\left(lay.di\right)}=0,45\left(mol\right)\)
=> \(V_{H_2\left(pứ\right)}=0,45\cdot22,4=10,08\left(l\right)\)
b) Theo pthh : \(n_{Fe\left(spu\right)}=\frac{2}{3}n_{H_2\left(pứ\right)}=0,3\left(mol\right)\)
=> \(m_{Fe}=16,8\left(g\right)\)
=> \(\hept{\begin{cases}\%m_{Fe}=\frac{16,8}{24,8}\cdot100\%\approx67,74\%\\\%m_{Fe_2O_3}\approx100\%-67,74\%=32,26\%\end{cases}}\)
c) Theo pthh : \(n_{Fe_2O_3\left(bi.khu\right)}=\frac{1}{3}n_{H_2\left(pứ\right)}=0,15\left(mol\right)\)
Mà thực tế, \(n_{Fe_2O_3}=\frac{32}{160}=0,2\left(mol\right)\)
=> \(H\%=\frac{0,15}{0,2}\cdot100\%=75\%\)