a) \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,2------->0,6---------->0,2
\(m_{Fe_2\left(SO_4\right)_3}=0,2.400=80\left(g\right)\)
b) \(m_{H_2SO_4}=0,6.98=58,8\left(g\right)\Rightarrow m_{dd.H_2SO_4}=\dfrac{58,8.100}{20}=294\left(g\right)\)