\(Fe_2O_3+H_2\rightarrow A+H_2O\)
\(\Rightarrow\) m rắn giảm=m O bị khử=32-27,2=4,8 gam
\(\rightarrow\) nO bị khử=\(=\frac{4,8}{16}=0,3\left(mol\right)\) \(\rightarrow\) nH2 =nO bị khử=0,3 mol
\(\rightarrow V_{H2}=V=0,3.22,4=6,72\left(l\right)\)
Ta có: \(n_{Fe2O3}=\frac{32}{60}=0,2\left(mol\right)\)
\(\rightarrow n_{Fe}=2n_{Fe2P3}=0,4\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,4.56=22,4\left(g\right)\Rightarrow\%m_{Fe_{trong.A}}=\frac{22,4}{27,2}=82,35\%\)