When ${\mathrm{FeCl}}_3$ is ignited in an atmosphere of pure oxygen, this reaction takes place :

$4{\mathrm{FeCl}}_3\left(s\right)+3{\mathrm{O}}_2\left(g\right)⟶2{\mathrm{Fe}}_2{\mathrm{O}}_3\left(s\right)+6{\mathrm{Cl}}_2\left(g\right)$

If 3 moles ${\mathrm{FeCl}}_3$ of are ignited in the presence of 2 moles of ${\mathrm{O}}_2$ gas, which of the following statements regarding to the given reaction is/are correct?

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:

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