解：設(shè)消耗氧氣的質(zhì)量為$y，$生成氧化鎂的質(zhì)量為$z。$
$2\text{Mg}+\text{O}_2\stackrel{\text{點(diǎn)燃}}{=}2\text{MgO}$
48 32 80
24g y z
$\frac{48}{32}=\frac{24\,\text{g}}{y}$
$y=\frac{24\,\text{g}\times32}{48}=16\,\text{g}$
$\frac{48}{80}=\frac{24\,\text{g}}{z}$
$z=\frac{24\,\text{g}\times80}{48}=40\,\text{g}$
答：需消耗氧氣$16\,\text{g}，$生成氧化鎂$40\,\text{g}。$
