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

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Answer 1 · 2017-09-15T15:47:17

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

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?

When a 325*g325*g mass of octane is combusted, what mass of water will result?