Question

What mass of water will result from complete combustion of a `216*g` mass of `"octane"`?

Answer 1

You got the stoichiometric equation....

Explanation:

...which we could represent as...

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

And thus per mole of octane, 9 moles of water should result upon complete combustion....

`"Moles of octane"=(216*g)/(114.23*g*mol^-1)=1.89*mol`

And given this molar quantity,

`1.89*molxx9xx18.01*g*mol^-1=306.5*g` of water will result....

All I am doing is following the stoichiometric equation....