Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PT: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
____0,1_______0,3________0,1 (mol)
a, \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4}{25\%}=117,6\left(g\right)\)
b, Ta có: m dd sau pư = 16 + 117,6 = 133,6 (g)
\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{133,6}.100\%\approx29,94\%\)