a) \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
b) \(n_{Fe_2O_3}=\dfrac{64}{160}=0,4\left(mol\right)\)
Theo PTHH: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,4\left(mol\right)\Rightarrow m_{Fe_2\left(SO_4\right)_3}=0,4.400=160\left(g\right)\)
c) Theo PTHH: \(n_{H_2SO_4}=3.n_{Fe_2O_3}=1,2\left(mol\right)\)
=> \(m_{H_2SO_{\text{4}}}=1,2.98=117,6\left(g\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{117,6}{180}.100\%=65,33\%\)