\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,05 0,15 0,05 ( mol )
\(m_{ddH_2SO_4}=\dfrac{0,15.98}{19,6\%}=75\left(g\right)\)
\(m_{ddspứ}=8+75=83\left(g\right)\)
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,05.400}{83}.100=24,09\%\)