Chemistry HELP!?! Please.?

Question

2.28 g of Octane is completely burned in 7.00 g of Oxygen...CHEMISTRY HELP!?

1) Which is the limiting reagent?
Why?

2) How much water vapor can be produced?
How?

Answer 1

First balance the reaction:

C8H18 + 25/2 O2 ---> 8 CO2 + 9 H2O

Or, if you prefer whole numbers:

2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O

2.28 g C8H18 / 114 g/mole = 0.020 moles
7.00 g O2 / 32 g/mole = 0.219 moles

You only have 10 times as many moles of oxygen as you do octane, and you need 12.5. So oxygen is your limiting reagent.

As for how much water you can produce:

0.219 moles O2 * 18 moles H2O/25 moles O2 * 18 g/mole = 2.84 g