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

1 Answer
Sep 15, 2017

Answer:

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?