A mixture of methane, propane and carbon monoxide contain 36.5% propane by volume. If its 200 mL are burnt in excess of air, the volume of CO₂ formed is:

Combustion of mixture takes place as

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

Let the volumes of CO, CH₄ and C₃H₈ in the mixture are a, b and c mL respectively, then a+b+c= 200  . . . . (i) 

and C = 73 $\left(2X36.5\right)$

Now from the combustion reactions, Volume of CO₂ formed

= a (from CO) + b (from CH₄) + 3c (from C₃H₈)

= (a + b) + 3c

= (200 - 73) + 3x73                  

= 127 + 219 = 346 mL