\( 1.0 \mathrm{~g} \) of \( \mathrm{Mg} \) is burnt with \( 0.28 \mathrm{~g} \) of \( \mathrm{O}_{2} \) in a closed vessel. Which reactant is left in excess and
how much?

For the reaction
\(2 \mathrm{Mg}+\mathrm{O}_{2} \rightarrow 2 \mathrm{M} \mathrm{gO}\)
\(48 \mathrm{~g}\) of \(\mathrm{Mg}\) requires \(32 \mathrm{~g}\) of \(\mathrm{O}_{2}\)
Therefore, \(1 \mathrm{~g}\) of \(\mathrm{Mg}\) require \(\frac{32}{48} \mathrm{~g}\) of \(\mathrm{O}_{2}=0.66 \mathrm{~g}\)
But available \(\mathrm{O}_{2}\) is \(0.28 \mathrm{~g}\)
So, \(\mathrm{O}_{2}\) is the limiting reagent and \(\mathrm{Mg}\) is the excess reagent.
\(32 \mathrm{~g}\) of \(\mathrm{O}_{2}\) needs \(48 \mathrm{~g}\) of \(\mathrm{Mg}\)
\(1 \mathrm{~g}\) of \(\mathrm{O}_{2}\) needs \(\frac{48}{32} \mathrm{~g}\) of \(\mathrm{Mg}\)
\(0.28\) of \(\mathrm{O}_{2}\) needs \(\frac{48}{32} \times 0.28=0.42 \mathrm{~g}\)
Mg available \(=1 \mathrm{~g}\)
Mg left \(=1-0.42=0.58 \mathrm{~g}\)