\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
-> ko giải phóng H2
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,15 0,15 ( mol )
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,15.56=8,4g\\m_{Fe_2O_3}=24,4-8,4=16g\end{matrix}\right.\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
0,15 <---------------------------- 0,15
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,15.22,4=8,4\left(g\right)\\m_{Fe_2O_3}=24,4-8,4=16\left(g\right)\end{matrix}\right.\)
