\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,1------>0,3------------>0,1
=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\Rightarrow m_{dd.H_2SO_4}=\dfrac{29,4.100}{20}=147\left(g\right)\)
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{16+147}.100\%=24,54\%\)