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CHEMICAL REACTIONS
 

CHEMICAL REACTIONS
by College_Student on 11/07/04 at 20:57:46

Hello:
The problem is the following:
Driving your car at 100 Miles at 20 mi/gal; assume gasoline is 100% octane; Density is 0.85g/cubic centimeter.
a) How many grs. of CO2 will you add to the atmosphere?
b) How many cubic meters of CO@ are generated? CO2 density is 1.98 Kg/cubic meter (gas at 298 Kelvin Degres)

My professor is kind of old and does not explain much, I am having lots of trouble understanding what to convert.  I think I have to convert 20 mi/gal to cubic cm but I am not sure...  Please help!!! :-[


Re: CHEMICAL REACTIONS
by Robert Fogt on 11/07/04 at 23:36:51

I wasn't able to find a direct answer to your question.

[url]http://www.turborick.com/gsxr1127/gasoline.html[/url]

[url]http://www.infinitepower.org/calc_carbon.htm[/url]

My only hope is you can find an email address at one of those two links who can answer your question. They talk directly about CO[sub]2[/sub] and automotive gasoline, so I assume one will be able to help you.


Re: CHEMICAL REACTIONS
by william on 11/14/04 at 13:38:17

If I understand the qustion you are travelling 100 miles so you use 5 gallons. What you need to do is write out the chemical reaction of the combustion of octane and then ballance that equation. In order to do the stoichiometry you need to convert from gallons of Octane to moles. Do this by converting gallons to cubic centimeters then to grams and use the molar mass of octane to determine the number of moles. After you know how much Octane is being burned and what the ratio of Octane to CO2 is just multiply by that ratio. Part b is just a simple conversion.


Re: CHEMICAL REACTIONS
by Albert on 12/02/04 at 14:13:11

100 miles at 20 mi/galon= 5 gallons of octane used

5 Gal of octane is 18,9 liters

18,9 liters is 18,9 l x 0,85 Kg/l = 16,1 kg octane

Octane is C8H18 = 114 gram/mole

1 mole of octane gives 8 moles of CO2 when burned.

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

16,1 kg/114g/mol = 141 mole octane

141 mole octane gives 8 x 141 = 1130 mol CO2

1 mole CO2 = 44 gram/mole

So 1130 mole CO2 = 1130 x 44 = 49720 gram= 49,720 Kg

So you added 49,72 Kg of CO2 to the air driving 100 miles

How much is that in cubic meters:

1 qM contains 1,98 Kg CO2
thus: 49,72/1,98 = 25,1 cubic meters of CO2 released to the air.

Easy isnt it?

;)


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