The combustion of petrol can be approximately represented by the

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The combustion of petrol is represented by the equation 2C8H18(g) + 25O2 → 16CO2 + 18H2O. The density of octane is 0.7 g/cm³, leading to a mass calculation for 1L of petrol. The user attempts to find the mass of oxygen consumed during combustion, initially calculating the amount of petrol in moles and then attempting to determine the corresponding oxygen mass. There is confusion regarding the stoichiometric ratio, specifically whether to multiply by 25 or 12.5 for the oxygen amount. The correct mass of oxygen used when burning 1L of petrol is approximately 2.5 x 10^3 grams, indicating a significant error in the initial calculations.
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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? I am 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
 
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Mass of 1L is not 0.0007g.

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