$M g+\frac{1}{2} O_{2} \rightarrow M g O$

Moles: $\frac{1.0}{24} ; \frac{0.56}{32}$

${\frac{0.5}{12} ; \frac{0.07}{4}}$

${\frac{0.5}{12}-x ; \frac{0.07}{4}-\frac{x}{2}}$

Oxygen is limiting reagent so, $\frac{0.07}{4}-\frac{x}{2}=0$

$x=\frac{0.07}{2}$

Excess $M g=\frac{0.5}{12}-\frac{0.07}{2} \;\mathrm{mol}$

Mass of $M g$ is $=1-0.7 \times 12=0.16\; \mathrm{g}$

Thus, when $1.0 \;g$ of magnesium is burnt with $0.56 \;\mathrm{g} O_{2}$ in a closed vessel, $0.16\; \mathrm{g}$ magnesium is left in excess.