\(\left\{{}\begin{matrix}n_{Mg}=0,05\left(mol\right)\\n_{H2SO4}=0,075\left(mol\right)\end{matrix}\right.\)
\(PTHH:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Ta thấy nMg < nH2SO4 nên H2SO4 dư
\(\Rightarrow n_{H2}=n_{Mg}=0,05\left(mol\right)\)
\(\Rightarrow V_{H2}=0,05.22,4=1,12\left(l\right)\)
\(\Rightarrow n_{MgSO4}=n_{Mg}=0,05\left(mol\right)\)
\(\Rightarrow m_{MgSO4}=0,05.120=6\left(g\right)\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2\)
\(n_{Fe2O3}=\frac{1}{60}\left(mol\right)\Rightarrow m_{Fe2O3}=160.\frac{1}{60}==2,67\left(g\right)\)