\(n_{Fe2O3}=\dfrac{16}{160}=0,1\left(mol\right)\)
a) Pt : \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,1-------->0,3------------>0,1
b) \(m_{ddH2SO4}=\dfrac{0,3.98}{9,8\%}.100\%=300\left(g\right)\)
c) \(m_{ddspu}=16+300=316\left(g\right)\)
\(C\%_{Fe2\left(SO4\right)3}=\dfrac{0,1.400}{316}.100\%=12,66\%\)