\(n_{H_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{Fe}=\frac{8,96}{56}=0,16\left(mol\right)\)
\(PTHH:3H_2+Fe_2O_3\rightarrow2Fe+3H_2O\)
Ta có:\(n_{Fe}=0,16\)
\(\rightarrow n_{H_2}=\frac{0,16.3}{2}=0,24\left(mol\right)\)
\(\rightarrow\) Hiệu suất phản ứng: \(H=\frac{0,24.100}{0,3}=80\%\)