You need to post the balanced equation; we will check it for you.

The equation will tell how much octane is consumed by 0.810 moles of O2. The difference is the moles or grams of residual octane.

Post your work and we will check it for you.

You need to post the balanced equation; we will check it for you.

The equation tells how many moles of water is produced from the 0.810 moles of O2.

Post your results and we will check that too.

Balanced Equation C8H18(g) + O2(g) → CO2(g) + H2O(g). 2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)

C8H18(g) + O2(g) → CO2(g) + H2O(g). 2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)

Ok, so, 2 moles of C8H18 consumes 25 moles of oxygen gas.

So, 1 mole of octane consumes (25/2) = 12.5 moles of oxygen gas.
or 1 mole of oxygen burns (2/25) = 0.08 moles of octane

And,

0.310 moles of octane consume (12.5x0.310) = 3.875 moles of oxygen gas.

Or

0.810 moles of oxygen gas burns (0.08x0.810) = 0.0648 moles of octane.

How much octane is left can be easily found now :)

From what I told you to do in my previous post, try working the amount of moles of water out now :)

Maybe this is a trick question: there is such a small amount of oxygen that I wonder if it will burn at all!