Ta có: \(n_{H_2SO_4}=\dfrac{25.19,6\%}{98}=0,05\left(mol\right)\)
PT: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
____1/60________0,05________1/60 (mol)
⇒ m dd sau pư = 1/60.160 + 25 = 83/3 (g)
\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{\dfrac{1}{60}.400}{\dfrac{83}{3}}.100\%\approx24,1\%\)