For the reaction    ${\mathrm{Fe}}_2{\mathrm{O}}_3+3\mathrm{CO}\to 2\mathrm{Fe}+3{\mathrm{CO}}_2,$  The volume of carbon monoxide required to reduce one mole of ferric oxide is 

$\begin{array}{c}{\mathrm{Fe}}_2{\mathrm{O}}_3+3\mathrm{CO}\to 2\mathrm{Fe}+3{\mathrm{CO}}_2\\ 1\mathrm{vol}.\\ 1\mathrm{mol}.3\mathrm{vol}.\\ (\because \mathrm{vol}\mathrm{\%}=\mathrm{mol}\mathrm{\%})\end{array}\begin{array}{r}2\mathrm{vol}.\\ 2\mathrm{mol}.\end{array}\begin{array}{r}3\mathrm{vol}.\\ 3\mathrm{mol}.\end{array}$

One gram mol of any gas occupies 22.4 litre at NTP. 1 mol of 

${\mathrm{Fe}}_2{\mathrm{O}}_3$  requires $3 \mathrm{mol}$ of CO for its reduction i, e $1\text{ mol of }{\mathrm{Fe}}_2{\mathrm{O}}_3$ required $3\times 22.4$ litre of $67.2{\mathrm{dm}}^3$  

CO to get itself reduced.