\(n_{Fe}=\dfrac{4,2}{56}=0,075mol\\ Fe_2O_3+3H_2\xrightarrow[]{t^0}2Fe+3H_2O\\ n_{Fe_2O_3}=\dfrac{0,075}{2}=0,0375mol\\ n_{H_2}=\dfrac{0,075.3}{2}=0,1125mol\\ m_{Fe_2O_3}=0,0375.160=6g\\ V_{H_2,đktc}=0,1125.22,4=2,52l\\ V_{H_2,đkc}=0,1125.24,79=2,788875l\)