# Question #934b0

Dec 16, 2016

We need (i) a stoichiometrically balanced equation:

${C}_{4} {H}_{10} + \frac{13}{2} {O}_{2} \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(g\right)$

#### Explanation:

And then (ii) equiv quantities of reactant and product:

$\text{Moles of butane}$ $=$ $\frac{5.80 \cdot g}{58.12 \cdot g \cdot m o {l}^{-} 1} = 0.0998 \cdot m o l$

$\text{Moles of dioxygen} = \frac{1 \cdot a t m \times 15.0 \cdot L}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 273.15 K} = 0.669 \cdot m o l$

(This is not quite $\text{STP}$, but you will have to adapt this expression.) At any rate there is stoichiometric dioxygen for complete combustion, which we must presume.

Given the equation (i), $4 \times 0.0998 \cdot m o l$ $C {O}_{2}$ are evolved per equiv butane.

$V = \frac{n R T}{P}$

$= \frac{4 \times 0.0998 \cdot m o l \times \frac{0.0821 \cdot L \cdot a t m}{K \cdot m o l} \times 273.15 K}{1 \cdot a t m} = 8.95 L$

And thus approx. $9 \cdot L$ carbon dioxide gas evolve from complete combustion.