The percentage by volume of C₃H₈ in a gaseous mixture of C₃H₈, CH₄. and CO is 20. When 100 mL of the mixture is burnt in excess of O₂, the volume of CO₂ produced is:

100mL gaseous mixture contain 20 mL C₃H₈
So, volume of CH₄ and CO
=(100 -  20) = 80 mL
$\begin{array}{l}{\mathrm{C}}_3{\mathrm{H}}_8+5{\mathrm{O}}_2⟶3{\mathrm{CO}}_2+4{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{CH}}_4+2{\mathrm{O}}_2⟶{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}\\ \mathrm{CO}+\frac{1}{2}{\mathrm{O}}_2⟶{\mathrm{CO}}_2\end{array}$
80 mL (CH₄ and CO) will produce 80 mL ${\mathrm{CO}}_2$
C₃H₈ will produce = 3 x 20 = 60 mL
Total CO₂ produce = 80 +60  = 140