# When a 325*g mass of octane is combusted, what mass of water will result?

Sep 15, 2017

Well, we needs a stoichiometrically balanced equation to represent the combustion.......

#### Explanation:

And so....

${C}_{8} {H}_{18} \left(l\right) + \frac{25}{2} {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 9 {H}_{2} O \left(l\right)$

Is this stoichiometrically balanced? Does $\text{garbage in equal garbage out?}$

And thus we know that the complete combustion on one mole of octane leads to the formation of nine moles of water.....

But we have a molar quantity with respect to octane of.....

$\frac{325 \cdot g}{114.23 \cdot g \cdot m o {l}^{-} 1} \equiv 2.85 \cdot m o l$

And thus we gets 2.85*molxx9-=??*mol with respect to water. And what is the mass of the water? And is this reaction endothermic?