(Sửa đề: $1,24g\to 12,4g$)
$A_2O+H_2O\to 2AOH$
Bảo toàn khối lượng:
$m_{H_2O}=m_{AOH}-m_{A_2O}=16-12,4=3,6g$
$\Rightarrow n_{H_2O}=\dfrac{3,6}{18}=0,2(mol)$
$\Rightarrow n_{A_2O}=n_{H_2O}=0,2(mol)$
$M_{A_2O}=\dfrac{12,4}{0,2}=62=2M_A+16$
$\Leftrightarrow M_A=23(Na)$
$\to$ oxit là $Na_2O$