The number of moles of Fe₂O₃ formed when 5.6 lit of O₂ reacts with 5.6g of Fe?

5.6 lit of O₂=$\frac{5.6}{22.4}=0.25\mathrm{moles}$ ; 5.6g of Fe = $\frac{5.6}{56}=0.1\mathrm{mol}$

$$\begin{gathered} 4\mathrm{Fe}+3{\mathrm{O}}_2\to 2{\mathrm{Fe}}_2{\mathrm{O}}_3 \\ 4\mathrm{mol} 3\mathrm{mol} 2\mathrm{mol} \\ ? 0.25 \\ {\mathrm{n}}_{\mathrm{Fe}}=\frac{0.25}{3}\times 4=0.33\mathrm{mol} \end{gathered}$$

But only 0.1mol 'Fe' is given, So 'Fe' is Limiting reagent.

The number of moles of Fe₂O₃ formed by 0.1mol of 'Fe' =$\frac{0.1}{4}\times 2=0.05\mathrm{moles}$