# The combustion of petrol can be approximately represented by the

1. Jun 13, 2010

### WowINeedHelp

The combustion of petrol can be approximately represented by the equation

2C8H18(g) + 25O2 -----> 16CO2 + 18H2O

The density of Ocatane is 0.7 gcm(cubed)

What mass of oxygen is used up when 1L of petrol is burnt in a car engine?

ATTEMPT

0.7 gcm$$^{}-3$$ = 0.0007 gL$$^{}-1$$
Therefore
density = mass/volume
0.0007 gL$$^{}-1$$ = mass / 1 L
0.0007g = mass

amount (of petrol) = mass/molar mass
= 0.0007/228
= 0.000003 mols

therefore...amount of oxygen is 25 times amount i got for petrol?
so
n(o) = 0.000003*25
then
0.000075 = mass/molar mass
0.000075 = mass/32
0.0024g = mass

Have i blown it somewhere? Im not sure about the mols thing i did when going from petrol to oxygen... I mean do you times it by 25 OR 12.5 because the mol ratio is 2petrol + 25Oxygen
i think times by 25 because when finding molar mass for petrol I multiplied everything by 2

btw the right answer is appparentlyyyy 2.5*10^3 grams, what did i do? ahh
cheers

2. Jun 14, 2010

### Staff: Mentor

Re: stoichiometry

Mass of 1L is not 0.0007g.

Please use [noparse]cm3[/noparse] tags for cm3 and [noparse]H2[/noparse] for H2.