Mole Concept and Stoichiometry — Question 15

Given, 10 litres of this mixture contains 60% propane and 40% butane. Hence, propane is 6 litres and butane is 4 litres

$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3\text{ vol.} \\ 2\text{C}_4\text{H}_{10} & + & 13\text{O}_2 & \longrightarrow & 8\text{CO}_2 & + & 10\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 13 \text{ vol.} & \longrightarrow & 8\text{ vol.} \\ \end{matrix}$

1 Vol. C₃H₈ produces carbon dioxide = 3 Vol

So, 6 litres C₃H₈ will produce carbon dioxide = 3 x 6 = 18 litres

2 Vol. C₄H₁₀ produces carbon dioxide = 8 Vol

So, 4 litres C₄H₁₀ will produce carbon dioxide = $\dfrac{8}{2}$ x 4 = 16 litres

Hence, 34 (i.e., 18 + 16) litres of CO₂ is produced.