The number of moles of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ formed when $5.6$ lit of $O_2$ reacts with $5.6 g$ of $\mathrm{Fe}$ ?

$5.6 \mathrm{g}$ of $\mathrm{Fe}=5.6/56=0.1 \mathrm{mol}$

$5.6$ lit of ${\mathrm{O}}_2=5.6/22.4=0.25 \mathrm{mol}$ (At STP 1 mole of $O_2=22.4 L$ )

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

4 moles of $\mathrm{Fe}$ reacts with 3 moles of $O_2$ to give 2 moles of ${\mathrm{Fe}}_2{\mathrm{O}}_3$

Ratio of ${\mathrm{O}}_2/\mathrm{Fe}=3/4$

the ratio of ${\mathrm{O}}_2/\mathrm{Fe}=\frac{0.25}{0.1}=2.5$

For, 4 moles $\mathrm{Fe}$ we get 2 moles of ${\mathrm{Fe}}_2{\mathrm{O}}_3$.

So, for $0.1$ mole we get $=\frac{2}{4}\times 0.1=0.05$ moles of ${\mathrm{Fe}}_2{\mathrm{O}}_3$