\(Fe_3O_4+4H_2SO_4-->FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\left(1\right)\)
0,2____________________0,2________0,2
\(Ba+2H_2O-->Ba\left(OH\right)_2+H_2O\left(2\right)\)
\(n_{Fe_3O_4}=\frac{46,4}{232}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=\frac{400.19,6}{100.98}=0,8\left(mol\right)\)
Theo (1):
\(n_{H_2SO_4}=4n_{Fe_3O_4}=4.0,2=0,8\left(mol\right)=>pứ\) vừa đủ
\(Ba\left(OH\right)_2+FeSO_4-->BaSO_4+Fe\left(OH\right)_2\left(3\right)\)
0,2_________0,2
\(n_{Ba}=\frac{27,4}{137}=0,2\left(mol\right)\)
=> \(Ba\) pứ vừa đủ với d2 FeSO4
=> \(m_{d^2sau}=46,4+400+27,4-0,2.233-0,2.90=409,2\left(g\right)\)
d2 sau : Fe2(SO4)3
=> \(C\%_{d^2sau}=\frac{0,2.400}{409,2}.100=19,55\%\)
Đề bạn nó sao sao ấy nhỉ?
\(PTHH:Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\)
\(n_{Fe_3O_4}=0,2\left(mol\right);n_{H_2SO_4}=0,8\left(mol\right)\)
\(TL:\frac{0,2}{1}=\frac{0,8}{4}\) → 2 chất hết
\(C\%_{ddspu}=\frac{0,2.\left(152+400\right)}{46,4+400}.100\%=24,73\left(\%\right)\)
