Ta có: \(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)
\(m_{H_2SO_4}=245.20\%=49\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PT: \(Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,5}{4}\), ta được H2SO4 dư.
Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe_2\left(SO_4\right)_3}=n_{Fe_3O_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(pư\right)}=4n_{Fe_3O_4}=0,4\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,1\left(mol\right)\)
Ta có: m dd sau pư = 23,2 + 245 = 268,2 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{0,1.152}{268,2}.100\%\approx5,67\%\\C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{268,2}.100\%\approx14,81\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1.98}{268,2}.100\%\approx3,65\%\end{matrix}\right.\)
Bạn tham khảo nhé!
