In the combustion of octane (C8H18), a complete combuation would be
written as: C8H18 + O2 ----> CO2 + H2O. How many grams of oxygen would
be required to comletely combust 100 grams of octane? (NOTE: The
chemical equation must be balanced to get the answer)

Hi!!

The balanced equation for the combustion of octane is: 
2 C8H18 + 25 O2 ---------> 16 CO2 + 18 H2O 


That means means that in this reaction you will need 25 moles of O2 to
perform a complete combustion of 2 moles of C8H18.

The atomic weights involved in this problem are:
H:  1.0 g
C: 12.0 g
O: 16.0 g

We have:
1 mol of O2 weight 2x16g = 32g, then 
25 moles of O2 weight 25x32g = 800g

1 mol of C8H18 weight 8x12g + 18x1g = 96g + 18g = 114g, then
2 moles of C8H18 weight 2x114g = 228g

At this point we know that we need 800g of O2 to perform a complete
combustion of 228g of C8H18, then if we only have 100g of C8H18 to
burn, we will need:
(100g x 800g)/ 228g = 350.88g of O2. 

For reference about balancing equations:
"Ten tips for balancing simple equations":
http://antoine.frostburg.edu/chem/senese/101/reactions/balancing.shtml


I hope this helps you.

Best regards.
livioflores-ga

Here's what I came up with:

2C8H18 + 25O2  yields  16CO2 + 18H2O.

Hope this helps

very good, thank you.

Thank you for the good rating and the generous tip, I hope that I can
help you in the future.

Regards.
livioflores-ga