Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
a, Theo PT: \(n_{H_2O}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{H_2O}=0,2.18=3,6\left(g\right)\)
b, \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{2}{15}.56=\dfrac{112}{15}\left(g\right)\)