\(n_{H_2SO_4\left(p\text{ư}\right)}=\dfrac{0,18.1}{120\%}=0,15\left(mol\right)\)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
Đặt CTHH của oxit sắt là \(Fe_xO_y\)
PTHH:
`2Al + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2`
`Fe_xO_y + yH_2SO_4 -> Fe_x(SO_4)_y + yH_2O`
Theo PT:
\(n_{Al}=\dfrac{2}{3}n_{H_2}=0,02\left(mol\right)\\ n_{H_2SO_4}=n_{H_2}+y.n_{Fe_xO_y}\Rightarrow n_{Fe_xO_y}=\dfrac{0,15-0,03}{y}=\dfrac{0,12}{y}\left(mol\right)\)
\(\Rightarrow M_{Fe_xO_y}=\dfrac{6,94-0,02.27}{\dfrac{0,12}{y}}=\dfrac{160}{3}y\left(g/mol\right)\)
`=> 56x + 16y = (160y)/3 `
`=> x/y = 2/3`
`=> Fe_2O_3`
Theo PT: \(n_{Al_2\left(SO_3\right)}=\dfrac{1}{2}n_{Al}=0,01\left(mol\right);n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=\dfrac{0,12}{3}=0,04\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,01.342=3,42\left(g\right)\\m_{Fe_2\left(SO_4\right)_3}=0,04.400=16\left(g\right)\end{matrix}\right.\)
