For this reaction: C8H18 (g) + 25O2 (g) → 16CO2 (g) + 18H2O (g). How many grams of CO2 are emitted when 5.0 x 102 g of pure octane, (C8H18), is combusted

Answer 1

Your equation is not balanced and you do not need it anyway. The correct balancing is put a 1 in front of the octane, use the C balance fo determine you need 8CO₂, and use an H balance to determine you need 9H₂O. The O balance tells you that you need 8(2) + 9 O or 25/2 O₂.. Oh, I see what happened, you forgot to have a 2 in front of the octane (Personally, I prefer the form with one mole in front as it makes the equation per one mole of fuel - if you look up heats of combustion, they are referring to this form of the equation).

Anyway, all of the carbon in the octane goes exclusively to CO₂, so you can just use grams of C as a tie-in:

500 g oct (96 g C/114 g oct)(44g CO₂/12 g C)

You can use the usual 3-step stoichiometry calculation and the balance equation:

500 g oct(1 mole/114 g oct)(8 mole CO₂/1 mole oct)(44g_//1mole CO₂)

Please consider a tutor. Take care.