# How do we represent the stoichiometric combustion of octane? How much carbon dioxide will result if 15.4*g of octane are combusted?

Mar 16, 2017

We have the stoichiometric equation:

${C}_{8} {H}_{18} + \frac{25}{2} {O}_{2} \rightarrow 8 C {O}_{2} \uparrow + 9 {H}_{2} O$

#### Explanation:

And the given stoichiometry assures us that 8 equiv of carbon dioxide will result for each equiv of octane combusted:

$\text{Moles of octane} = \frac{15.4 \cdot g}{114.23 \cdot g \cdot m o {l}^{-} 1} = 0.135 \cdot m o l$.

And thus $0.135 \times 8$ $\text{mol}$ $C {O}_{2}$ will be evolved, and this represents a mass of :

$0.135 \cdot m o l \times 8 \times 44.01 \cdot g \cdot m o {l}^{-} 1 = 47.5 \cdot g$ $\text{carbon dioxide}$

How much water will be evolved. Will the mass of products equal the mass of reactants? Why?