Question

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

Answer 1

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

Explanation:

And so....

`C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) +9H_2O(l)`

Is this stoichiometrically balanced? Does `"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.....

`(325*g)/(114.23*g*mol^-1)-=2.85*mol`

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?