(Sửa kim loại thành oxit)
a,
$R_2O_x+ 2xHCl\to 2RCl_x+ xH_2O$
$n_{R_2O_x}=\dfrac{16}{2R+16x}$
$\Rightarrow n_{RCl_x}=\dfrac{32}{2R+16x}$
$\Rightarrow \dfrac{32}{2R+16x}=\dfrac{32,5}{R+35,5x}$
$\Rightarrow 32,5(2R+16x)=32(R+35,5x)$
$\Leftrightarrow R=\dfrac{56x}{3}$
$\Rightarrow x=3, R=56(Fe)$
b,
$n_{Fe_2O_3}=\dfrac{16}{160}=0,1 mol$
$\Rightarrow n_{HCl}=0,1.2x=0,6 mol$
$C_{M_{HCl}}=\dfrac{0,6}{0,12}=5M$