Mỗi phần có $5,92:2=2,96g$ A, gồm x mol Fe và y mol R.
- P1:
$n_{H_2}=\dfrac{1,568}{22,4}=0,07(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$2R+2nHCl\to 2RCl_n+nH_2$
$\Rightarrow x+0,5ny=0,07$ (1)
- P2:
$n_{NO}=\dfrac{1,344}{22,4}=0,06(mol)$
$Fe+4HNO_3\to Fe(NO_3)_3+NO+2H_2O$
$3R+4nHNO_3\to 3R(NO_3)_n+nNO+2nH_2O$
$\Rightarrow x+\dfrac{1}{3}ny=0,06$ (2)
$(1)(2)\Rightarrow x=0,04; ny=0,06$
$\Leftrightarrow y=\dfrac{0,06}{n}$
Ta có: $56.0,04+\dfrac{0,06R}{n}=2,96$
$\Leftrightarrow R=12n$
$n=2\Rightarrow R=24(Mg)$
$\to R$ là magie