a) Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_{H_2SO_4}=0,5x\left(mol\right)\\ n_{KOH}=0,45.2=0,9\left(mol\right)\)
PTHH:
Fe + H2SO4(l) ---> FeSO4 + H2
a----->a------------>a----------->a
2Al + 3H2SO4(l) ---> Al2(SO4)3 + 3H2
b----->1,5b------------>0,5b-------->1,5b
FeSO4 + 2KOH ---> Fe(OH)2↓ + K2SO4
a---------->2a-------->a
Al2(SO4)3 + 6KOH ---> 2Al(OH)3↓ + 3K2SO4
0,5b------------->3b---------->2b
4Fe(OH)2 + O2 --to--> 2Fe2O3 + 4H2O
a---------------------------->0,5a
2Al(OH)3 --to--> Al2O3 + 3H2O
2b----------------->b
Hệ pt \(\left\{{}\begin{matrix}56a+27b=11\\a+1,5b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)
b) nH2SO4 = nH2 = 0,4 (mol)
=> \(x=\dfrac{0,4}{0,5}=0,8M\)
Xét 2.0,1 + 0,5.6.0,2 = 0,8 (mol)
=> KOH dư
=> \(\left\{{}\begin{matrix}m_{Fe_2O_3}=0,5.0,1.160=8\left(g\right)\\m_{Al_2O_3}=0,2.102=20,4\left(g\right)\end{matrix}\right.\)
=> y = 8 + 20,4 = 28,4 (g)
