Gọi CT oxit là $M_2O_n$$M2On$ + 2n$HCl$ \rightarrow 2$MCl_n$ +n $H2O$
Mol oxit= 15,3/(2M+16n)
Mol muối=20,8/(M+35,5n)
Từ pt=>15,3.2/(2M+16n)=20,8/(M+35,5n)
=>30,6(M+35,5n)=20,8(2M+16n)=>11M=753,5n=>M=68,5n=>n=2 M=137 $Ba$
Mol $BaO$=0,1 mol=>mol $HCl$=0,2 mol=>m$HCl$=7,3g=>mdd=40g