The number of moles of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ formed when 5.6 lit of ${\mathrm{O}}_2$reacts with 5.6g of Fe is

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

$\frac{3}{2}mole O_2=2 mole Fe$

$\frac{3}{2}\times 22.4Lt of O_2=2\times 50=112 gram Fe$

$56 \mathrm{Lt} \mathrm{of} {\mathrm{O}}_2=\frac{112}{3\times 11.2}\times 5.6=\frac{112}{6}=18.66 \mathrm{gram} \mathrm{Fe}$

So, in this reaction Lt agent is iron, which determines amount of product.

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

1 mole Fe = 0.5 mole ${\mathrm{Fe}}_2{\mathrm{O}}_3$

0.1 mole Fe = 0.05 mole ${\mathrm{Fe}}_2{\mathrm{O}}_3$