Lộn nữa :vv

\(\%O=38,67\%\\ Ba_xH_yS_zO_t\\ x:y:z:t=\frac{41,39}{137} : \frac{0,6}{1} : \frac{19,34}{32} : \frac{38,67}{16}\\ x:y:z:t=0,3:0,6:0,6:2,4\\ x:y:z:t=1:2:2:8\\ \to BaSHO_4\)

À r e nhầm :v

\(n_{O_2}=0,0625(mol)\\ N=0,0625.6,023.10^{23}=3,76.10^{22}\)

\(Al_2O_x\\ M=27.2+16.x=102\\ \to x=3\\ \to Al_2O_3\)

\(\%_O=100\%-41,39\%-0,6\%-19,34\%=38,67\%\\ Ba_xH_yS_zO_t\\ x:y:z:t=\frac{41,39}{137} : \frac{0,6}{1} : \frac{19,34}{32} : \frac{38,67}{32}\\ x:y:z:t=0,3 : 0,6 : 0,6 : 1,2\\ x:y:z:t=1 : 2 : 2 : 4\\ \)

\(\text{Cho 3 kim loại vào } H_2O:\\ \text{- Tan: Ca}\\ \text{- Không tan: Mg; Ag}\\ Ca+2H_2O \to Ca(OH)_2+H_2\\ \text{Cho HCl vào 2 kim loại còn lại: }\\ \text{- Tan: Mg}\\ \text{- Không tan: Ag}\\ Mg+2HCl \to MgCl_2+H_2 \)

\(a/ BaO; SO_2\\ BaO+H_2O \to Ba(OH)_2\\ SO_2 +H_2O \to H_2SO_3\\ b/ SO_2\\ 2KOH+SO_2 \to K_2SO_3+H_2O\\ c/ BaO; Fe_2O_3\\ BaO+2HCl \to BaCl_2+H_2O\\\ Fe_2O_3+6HCl \to 2FeCl_3+3H_2O\)