Gọi x; y là số mol Fe, X.
- P2:
$n_{NO}=\dfrac{0,896}{22,4}=0,04(mol)$
$Fe+4HNO_3\to Fe(NO_3)_3+NO+2H_2O$
$3X+4nHNO_3\to 3X(NO_3)_n+nNO+2nH_2O$
$\Rightarrow x+\dfrac{1}{3}ny=0,04$ (1)
- P1:
$n_{H_2}=0,0475(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$2X+2nHCl\to 2XCl_n+nH_2$
$\Rightarrow x+0,5ny=0,0475$ (2)
(1)(2)$\Rightarrow x=0,025; ny=0,045$
$\Rightarrow y=\dfrac{0,045}{n}$
$\Rightarrow 56.0,025+X.\dfrac{0,045}{n}=1,805$
$\Leftrightarrow X=9n$
$n=9\Rightarrow X=27(Al)$