Gọi oxit KL là MO
\(MO+2HCl-->MCl2+H2O\)
\(n_{MO}=\frac{1}{2}n_{HC_{ }l}=0,1\left(mol\right)\)
\(M_{MO}=\frac{8}{0,1}=80\left(\frac{g}{mol}\right)\)
\(\Rightarrow M+16=80\Rightarrow M=64\left(Cu\right)\)
Vậy CTHH: CuO
Gọi oxit là RO
RO + 2HCl ---> RCl2+ H2
\(\frac{8}{M_R+16}\)-->0,2 (mol)
<=> \(\frac{8}{M_R+16}\)= 0,1 => MR= 64
=> R là Cu