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 20mL C₃​H_8​ 

So, volume of CH_4​ and CO =(100−20)=80mL 

$$\begin{gathered} {\mathrm{C}}_3{\mathrm{H}}_8+5{\mathrm{O}}_2 \to 3{\mathrm{CO}}_2+4{\mathrm{H}}_2\mathrm{O} \\ {\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O} \\ \mathrm{CO}+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{CO}}_2 \end{gathered}$$

80mL (CH₄​ and CO) will produce 80mLCO₂​ ; C₃​H_8​ will produce =3×20=60mL 

Total CO_2​ produced =80+60=140