Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)
a. PTHH: Fe3O4 + 4H2SO4 ---> FeSO4 + Fe2(SO4)3 + 4H2O
Theo PT: \(n_{H_2SO_4}=4.n_{Fe_3O_4}=4.0,01=0,04\left(mol\right)\)
=> \(m_{H_2SO_4}=0,04.98=3,92\left(g\right)\)
Theo đề, ta có: \(C_{\%_{H_2SO_4}}=\dfrac{3,92}{m_{dd_{H_2SO_4}}}.100\%=20\%\)
=> \(m_{dd_{H_2SO_4}}=19,6\left(g\right)\)
b. Ta có: \(m_{dd_{SauPỨ}}=2,32+19,6=21,92\left(g\right)\)
Theo PT: \(n_{FeSO_4}=n_{Fe_2\left(SO_4\right)_3}=n_{Fe_3O_4}=0,01\left(mol\right)\)
=> \(m_{FeSO_4}=0,01.152=1,52\left(g\right)\)
\(m_{Fe_2\left(SO_4\right)_3}=0,01.400=4\left(g\right)\)
=> \(m_{SauPỨ}=1,52+4=5,52\left(g\right)\)
=> \(C_{\%_{SauPỨ}}=\dfrac{5,52}{21,92}.100\%=25,18\%\)
