Bạn tham khảo:
$a/$
Gọi:
$n_{Fe}=a(mol)$
$n_R=b(mol)$
$\to m_{hh}=56.2a+2R.b=7,22(1)$
P1:
$Fe+2HCl \to FeCl_2+H_2$
$2R+2nHCl \to 2RCl_n+nH_2$
$n_{H_2}=0,095(mol)$
$\to n_{H_2}=a+\frac{n}{2}b=0,095(2)$
P2:
$Fe+4HNO_3 \to Fe(NO_3)_3+NO+2H_2O$
$3R+4nHNO_3 \to 3R(NO_3)_n+nNO+4nH_2O$
$n_{NO}≈0,08(mol)$
$n_{NO}=a+\frac{n}{3}b=0,08(3)$
$(1)(2)(3)$
$a=0,05$
$R.b=0,81$
$n.b=0,09$
$M_R=\frac{0,81.n}{0,09}=9n(g/mol)$
$n=3; R=27$
$\to R: Al$
$b/$
$\%m_{Fe}=\frac{2.0,05.56}{7,22}.100\%=77,56\%$
$\%m_{Al}=22,44\%$