Gọi \(\left\{{}\begin{matrix}n_{Fe_3O_4}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\Rightarrow232a+27b=93,9\left(1\right)\)
Vì chất rắn B sau phản ứng + dd NaOH -> H2 => Al dư
B gồm: Al, Al2O3, Fe
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: \(3Fe_3O_4+8Al\xrightarrow[]{t^o}9Fe+4Al_2O_3\)
a-------->\(\dfrac{8}{3}a\)----->3------>\(\dfrac{4}{3}a\)
\(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)
0,1<-------0,1<----------------0,1<----------0,15
`=>` \(b-\dfrac{8}{3}a=0,1\left(2\right)\)
Từ `(1), (2) =>` \(\left\{{}\begin{matrix}a=0,3\left(mol\right)\\b=0,9\left(mol\right)\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}m_{Fe_3O_4}=0,3.232=69,6\left(g\right)\\m_{Al}=0,9.27=24,3\left(g\right)\end{matrix}\right.\)
Theo PT: \(n_{NaAlO_2}=n_{NaOH}=0,45.2=0,9\left(mol\right)\)
b) \(n_{Al\left(OH\right)_3}=\dfrac{62,4}{78}=0,8\left(mol\right)\)
PTHH: \(NaAlO_2+HCl+H_2O\rightarrow Al\left(OH\right)_3\downarrow+NaCl\)
`-` TH1: Kết tủa chưa bị hòa tan
Theo PT: \(n_{HCl}=n_{Al\left(OH\right)_3}=0,8\left(mol\right)\)
`=>` \(V=\dfrac{0,8}{2}=0,4\left(l\right)=400\left(ml\right)\)
`-` TH2: Kết tủa tan ra 1 phần
PTHH: \(Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\)
BTNT Na: `n_{NaCl} = n_{NaAlO_2} = 0,9 (mol)`
BTNT Al: `n_{AlCl_3} = n_{NaAlO_2} - n_{Al(OH)_3} = 0,1 (mol)`
BTNT Cl: `n_{HCl} = n_{NaCl} + 3n_{AlCl_3} = 1,2 (mol)`
`=>` \(V=\dfrac{1,2}{2}=0,6\left(l\right)=600\left(ml\right)\)
