1 g of Mg is burnt in a vessel containing 0.5 g of O_2. The reactant remaining unreacted is

Balanced chemical reaction is:

$2\mathrm{Mg}+{\mathrm{O}}_2\to 2\mathrm{MgO}$

According to the given reaction, 2 moles (or 48 g) of Mg needs one mole of oxygen (or 32 g) of oxygen to form one mole of MgO.

Given mass of Mg is 1 g and the mass of oxygen requires to react with 1 g of Mg is:

$\begin{array}{ccc}48 \mathrm{g} \mathrm{Mg}& =& 32 \mathrm{g} {\mathrm{O}}_2\\ 1 \mathrm{g} \mathrm{of} \mathrm{Mg}& =& 1 \mathrm{g} \mathrm{of} \mathrm{Mg}\times \frac{32 \mathrm{g} \mathrm{of} {\mathrm{O}}_2}{48 \mathrm{g} \mathrm{of} \mathrm{Mg}}\\ & =& 0.66 \mathrm{g} \mathrm{of} {\mathrm{O}}_2\end{array}$

Given mass of oxygen is 0.5 g, therefore oxygen is the limiting reagent.

Mass of Mg requires to react with 0.5 g of oxygen is:

$\begin{array}{ccc}32 \mathrm{g} {\mathrm{O}}_2& =& 48 \mathrm{g} \mathrm{of} \mathrm{Mg}\\ 0.5 \mathrm{g} \mathrm{of} {\mathrm{O}}_2& =& 0.5 \mathrm{g} \mathrm{of} {\mathrm{O}}_2\times \frac{48 \mathrm{g} \mathrm{of} \mathrm{Mg}}{32 \mathrm{g} \mathrm{of} {\mathrm{O}}_2}\\ & =& 0.75 \mathrm{g} \mathrm{of} \mathrm{Mg}\end{array}$

$\begin{array}{ccc}\mathrm{Excess} \mathrm{amount} \mathrm{of} \mathrm{Mg}& =& \mathrm{Total} \mathrm{amount}-\mathrm{Used}\\ & =& 1 \mathrm{g}-0.75 \mathrm{g}\\ & =& 0.25 \mathrm{g}\end{array}$