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April 1, 2015

April 1, 2015

Posted by **stephanie** on Monday, February 26, 2007 at 11:04pm.

Chaum drives a car 10,000 miles during the year and the car gets 22 miles/gallon. How many pounds of carbon dioxide in a year will Chaum produce? The density of gasoline is 0.703 g/mL. The unbalanced reaction of gasonline combustion is:

C8H18 + O2 ¡ú CO2 + H2O

Balance the equation.

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

Chaum drove 10,000 miles in a year. How many gallons of gasoline did Chaum use?

10,000 miles x (1 mile/22 gallons) = ??gallons in one year.

Convert gallons/year to liters/year and convert that to grams/year using the density.

Then use ordinary stoichiometry with the equation to determine the grams of CO2 produced by the grams of C8H18 used.

Convert grams to lbs. There are 454 (approximately) grams/lb.

Post your work if you get stuck.

i got an answer, but i don't know if it's right...i got 454.454 gal/yr. then i converted the gal/yr to ml/yr and got 1.72 X 10^6 mL. and from there I ended up getting 7520 lb/yr. is that answer correct?

I am with you to 1.72*10^6 mL. But I don't get 7520 lb/yr.

Check my numbers. Converting 1.72*10^6 mL to grams (@0.703 g/mL)=1.209^10^6 grams. Then through the stoichiometry to obtain 3.726*10^6 g CO2 which converts to 8,215 lbs CO2 if I didn't err. This is such a long problem it is easy to make a small error and never spot it. Post your work and I can find the difference between our answers. Thanks for using Jiskha.

ok, so this is how i got 7520 lb/yr:

(1.72*10^6 mL gas)(0.703 g gas/ 1 mL gas)(1 mole gas/130.26 g gas)(16 mole CO2/ 2 mole gas)(46.01 g CO2/1 mole CO2)(1 kg/1000g)(2.20lb/1kg)= 7520 lb/yr

It appears you have used 130.26 for the molar mass of C8H18. If so, that is part of the trouble you are having.

8x12 = about 96

18x1 = about 18

96+18 = about 114.

You need to confirm this but I added C8H18 up to be 114.23.

ALSO, I note you used 46.01 for the molar mass of CO2. It should be 44.01. I think those two corrections will get and answer close to 8215 lbs CO2.

oh...yeah ok i see that now. ya so i got close to your answer now. Thank you very much for your help!!! :)

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