Gọi 0,5x; 0,5y là số mol Fe, R mỗi phần.
- P2:
$n_{NO}=\dfrac{3,136}{22,4}=0,14(mol)$
$Fe+4HNO_3\to Fe(NO_3)_3+NO+2H_2O$
$3R+4nHNO_3\to 3R(NO_3)_n+nNO+2nH_2O$
$\Rightarrow 0,5x+\dfrac{0,5}{3}ny=0,14$ (1)
- P2:
$n_{H_2}=0,1(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$2R+2nHCl\to 2RCl_n+nH_2$ (Nếu $R>H$)
+ TH1: $R>H$: $0,5x+0,25ny=0,1$ (2)
(1)(2)$\Rightarrow $ nghiệm âm, loại
+ TH2: $R<H$: $0,5x=0,1$ (3)
(1)(3)$\Rightarrow x=0,2; ny=0,24$
$\Rightarrow y=\dfrac{0,24}{n}$
Ta có:
$56.0,2+\dfrac{0,24R}{n}=18,88$
$\Leftrightarrow R=32n$
$n=2\to R=64(Cu)$