For the provided reaction,

$3\text{ moles }$of O₂ requires 4 moles of ${\mathrm{FeCl}}_3$

Thus, O₂ is a limiting reagent.

Thus, 2 mole of O₂ will require:

 $\frac{2\times 4}{3}=2.67\text{ moles }$of ${\mathrm{FeCl}}_3$

Thus, the excess of ${\mathrm{FeCl}}_3$ is $3-2.67=0.33\mathrm{mol}$

Moles of Fe₂O₃ produced:

$\mathrm{Moles} \mathrm{of} {\mathrm{Fe}}_2{\mathrm{O}}_3=\frac{2 \mathrm{moles} \mathrm{of} {\mathrm{Fe}}_2{\mathrm{O}}_3}{3 \mathrm{moles} {\mathrm{O}}_2}\times 2 \mathrm{moles} {\mathrm{O}}_2=1.33 \mathrm{moles} {\mathrm{Fe}}_2{\mathrm{O}}_3$