a)
\(3Fe_xO_y+2yAl\underrightarrow{t^o}3xFe+yAl_2O_3\) (1)
\(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\) (2)
\(2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\) (3)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\) (4)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (5)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (6)
b)
Y chứa Al, Fe, Al2O3
- Xét phần 1: \(\left(n_{Al};n_{Fe};n_{Al_2O_3}\right)=\left(a;b;c\right)\)
\(n_{H_2}=\dfrac{0,168}{22,4}=0,0075\left(mol\right)\)
nNaOH = 0,125.0,2 = 0,025 (mol)
PTHH: \(2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)
0,005<----0,005<------------------------0,0075
\(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
0,01<-----0,02
=> a = 0,005 (mol); c = 0,01 (mol)
- Xét phần 2: \(\left(n_{Al};n_{Fe};n_{Al_2O_3}\right)=\left(ak;bk;ck\right)\)
\(n_{H_2SO_4}=\dfrac{2,106}{22,4}=0,09\left(mol\right)\)
Theo PTHH (5), (6) => 1,5ak + bk = 0,09
=> \(b=\dfrac{0,09-0,0075k}{k}\) (*)
Ta có: mY = mX
=> 27(a + ak) + 56(b + bk) + 102(c + ck) = 9,66
=> \(\left(k+1\right)\left(56b+1,155\right)=9,66\) (**)
(*)(**) => \(\left[{}\begin{matrix}k=\dfrac{16}{7}\\k=3\end{matrix}\right.\)
TH1: \(k=\dfrac{16}{7}\) => \(\left\{{}\begin{matrix}a=0,005\left(mol\right)\\b=0,031875\left(mol\right)\\c=0,01\left(mol\right)\end{matrix}\right.\)
Theo PTHH (1): \(\dfrac{n_{Fe}}{n_{Al_2O_3}}=\dfrac{3x}{y}=\dfrac{0,031875}{0,01}\Rightarrow\dfrac{x}{y}=\dfrac{17}{16}\) (L)
TH2: k = 3 => \(\left\{{}\begin{matrix}a=0,005\left(mol\right)\\b=0,0225\left(mol\right)\\c=0,01\left(mol\right)\end{matrix}\right.\)
Theo PTHH (1): \(\dfrac{n_{Fe}}{n_{Al_2O_3}}=\dfrac{3x}{y}=\dfrac{0,0225}{0,01}\Rightarrow\dfrac{x}{y}=\dfrac{3}{4}\)
=> CTHH: Fe3O4
c) 9,66 (g) X chứa \(\left\{{}\begin{matrix}Fe_3O_4:0,03\left(mol\right)\\Al:0,1\left(mol\right)\end{matrix}\right.\)
=> 4,83 (g) X chứa \(\left\{{}\begin{matrix}Fe_3O_4:0,015\left(mol\right)\\Al:0,05\left(mol\right)\end{matrix}\right.\)
PTHH: \(2Fe_3O_4+10H_2SO_{4\left(đ,n\right)}\rightarrow3Fe_2\left(SO_4\right)_3+SO_2+10H_2O\)
0,015----------------------------------->0,0075
\(2Al+6H_2SO_{4\left(đ,n\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)
0,05--------------------------------->0,075
=> V = (0,0075 + 0,075).22,4 = 1,848 (l)
