\(n_{Fe}=\dfrac{6,72}{56}=0,12\left(mol\right)\\ Fe+H_2SO_{4\left(loãng\right)}\rightarrow FeSO_4+H_2\uparrow\\ Mol:0,12\rightarrow0,12\rightarrow0,12\rightarrow0,12\\ V_{H_2}=0,12.22,4=2,688\left(l\right)\\ m_{FeSO_4}=0,12.152=18,24\left(g\right)\\ PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ Mol:0,04\leftarrow0,12\rightarrow0,08\\ m_{Fe}=0,08.56=4,48\left(g\right)\)