1 g of Mg is burnt in a closed vessel containing 0.5 g of O₂. which reactant is limiting reagent and how much of the excess reactant will be left?

$\underset{2\mathrm{x}24}{2\mathrm{Mg}}+\underset{2\mathrm{x}16}{{\mathrm{O}}_2}\to \underset{2(24+16)}{2\mathrm{MgO}}$
48 g of Mg requires 32 g of O₂
1 g of Mg requires $\frac{32}{48}=0.66\mathrm{g} \mathrm{of} {\mathrm{O}}_2$
Oxygen available = 0.5 g
Hence, O₂ is limiting reagent.
32 g of O₂ reacts with 48 g of Mg
0.5 g of O₂ will react with $\frac{48}{32}\times 0.5=0.75 \mathrm{g} \mathrm{of} \mathrm{Mg}$
Excess of Mg = (1.0 - 0.75) = 0.25 g