PTHH: \(3Fe_3O_4+8Al\underrightarrow{t^o}9Fe+4Al_2O_3\)
=> \(\dfrac{n_{Fe}}{n_{Al_2O_3}}=\dfrac{9}{4}\)
P1: Gọi \(\left(n_{Fe};n_{Al_2O_3};n_{Al}\right)=\left(a;b;c\right)\)
=> \(\dfrac{a}{b}=\dfrac{9}{4}\) (1)
\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
PTHH: \(2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)
0,04<--------------------------------------0,06
=> c = 0,04 (mol) (2)
P2: Gọi \(\left(n_{Fe};n_{Al_2O_3};n_{Al}\right)=\left(ak;bk;ck\right)\)
\(n_{H_2}=\dfrac{14,112}{22,4}=0,63\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
ak----------------------->ak
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
ck------------------------>1,5ck
=> ak + 1,5ck = 0,63 (3)
Ta có: mX = mY = 56(a+ak) + 102(b+bk) + 27(c+ck)
=> (56a + 102b + 27c)(k+1) = 93,9 (4)
(1)(2)(3)(4) => a = 0,36 (mol); b = 0,16 (mol); c = 0,04 (mol); k = 1,5
Bảo toàn Fe: \(n_{Fe_3O_4}=0,3\left(mol\right)\)
Bảo toàn Al: \(n_{Al}=0,9\left(mol\right)\)
\(\left\{{}\begin{matrix}m_{Fe_3O_4}=0,3.232=69,6\left(g\right)\\m_{Al}=0,9.27=24,3\left(g\right)\end{matrix}\right.\)z
