27 gm of Al will react completely with 

$4\mathrm{Al}+3{\mathrm{O}}_2⟶2{\mathrm{Al}}_2{\mathrm{O}}_3$

$2\mathrm{Al}+\frac{3}{2}{\mathrm{O}}_2⟶{\mathrm{Al}}_2{\mathrm{O}}_3$

Moles of $\mathrm{Al}=\frac{27}{27}=1\mathrm{mol}$

2 mole Al gives $\frac{3}{2}\text{ mole }{\mathrm{O}}_2$

1 mole $\mathrm{Al}$ gives $\frac{3}{4} \mathrm{mole} {\mathrm{O}}_2=0.75 \mathrm{mole} {\mathrm{O}}_2$

Mass of oxygen $=\frac{3}{4}\times 32=24\text{ gram }$.

Molecules of O₂ is calculated as:

$\mathrm{Molecules}=0.75{\mathrm{N}}_{\mathrm{A}}(\mathrm{Avogardo}\text{'}\mathrm{s} \mathrm{number})$

Volume occupies by one mole of oxygen at 1 atm and 273 K is 22.4 L. Volume occupies by 0.75 mole of oxygen:

$\mathrm{Volume}=\frac{22.4 \mathrm{L}}{1 \mathrm{mol}}\times 0.75 \mathrm{mol}=16.8 \mathrm{L}$