The number of moles of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ formed when 0.5 moles of ${\mathrm{O}}_2$ and 0.5 moles of Fe are allowed to react are

$2\mathrm{Fe}+\frac{3}{2}{\mathrm{O}}_2\to {\mathrm{Fe}}_2{\mathrm{O}}_3$

2mole Fe = 1.5 mole ${\mathrm{O}}_2=1 \mathrm{mole} {\mathrm{Fe}}_2{\mathrm{O}}_3$

0.5 mole Fe = $\frac{1.5}{2}\times 0.5=\frac{1.5}{2}\times \frac{1}{2}=\frac{1.5}{4}=0_2$

So, here limiting reagent is Fe

2 mole Fe = 1 mole ${\mathrm{Fe}}_2{\mathrm{O}}_3$

$4 \mathrm{Fe}+3{\mathrm{O}}_2\to 2{\mathrm{Fe}}_2{\mathrm{O}}_3$

4               3                 2

$=\frac{0.5\times 2}{4}$

= 0.25