Giả sử $m_1=30g$
$\Rightarrow m_{\text{oxit}}=50g$
Bảo toàn khối lượng:
$m_{O_2}=50-30=20g$
$\Rightarrow n_{O_2}=\dfrac{20}{32}=0,625 mol$
$2xM+yO_2\to 2M_xO_y$
$\Rightarrow n_M=\dfrac{1,25x}{y}$
$\Rightarrow M_M=\dfrac{30y}{1,25x}=\dfrac{24y}{x}$
$\Rightarrow x=y=1; M_M=24(Mg)$