\(m_{Fe_2O_3\left(nguyên,chất\right)}=20.80\%=16g\)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,1 0,3 0,3 ( mol )
\(m_{H_2O}=0,3.18=5,4g\)
\(V_{H_2}=0,3.22,4=6,72l\)
\(n_{Fe_2O_3}=\dfrac{20}{160}.80\%=0,1\left(mol\right)\\
pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
0,1 0,3 0,3
=> \(m_{H_2O}=0,3.18=5,4\left(g\right)\\
V_{H_2}=0,3.22,4=6,72\left(l\right)\)