M2On+nCO\(\rightarrow\)2M+nCO2
CO2+Ba(OH)2\(\rightarrow\)BaCO3+H2O
nCO2=nBaCO3=\(\frac{19,7}{197}\)=0,1(mol)
\(\rightarrow\)nM2On=\(\frac{0,1}{n}\)
\(\rightarrow\)MM2On=232n
Ta có
2M+16n=232n\(\rightarrow\)M=108n
\(\rightarrow\)n=1 M=108
\(\rightarrow\)M là Bạc(Ag)
CT oxit Ag2O