2gm of Mg is burnt in a vessel containing 1.0gm of oxygen. The reactant remaining unreacted is

$Mg+\frac{1}{2}{O}_2\stackrel{}{\to }MgO$

               Limiting reagent = $O_2$

               $16gm{O}_2\stackrel{}{\to }24gmMg$

               $1gm{O}_2\stackrel{}{\to }?$

               $=\frac{24}{16}=\frac{3}{2}=1.5gmMg$

                Wt of ‘Mg’ Left = 2.0-1.5 = 0.5gm