# What volume of oxygen gas is needed to completely combust 0.202 L of butane gas?

${C}_{4} {H}_{10} \left(g\right) + \frac{13}{2} {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(l\right)$
The stoichiometrically balanced equation tells me unequivocally that for each equiv butane, I NEED $\frac{13}{2}$ equivs of dioxygen gas for complete combustion. Agreed?
So, since gaseous volume is proportional to the number of moles, at given temperature and pressure, we need a volume of dioxygen gas $= 0.202 \cdot L \times \frac{13}{2} = \text{a bit under 1.5 litres}$. How many litres of carbon dioxide gas will result?