The balanced equation minus the phases to save typing space.
2C8H18 + 25O2 ==> 16CO2+ 18H2O
21.4 mpg x #gallons = 1280 miles
Solve for gallons needed.
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Convert gallons octane to L octane.
Using density convert L octane to grams octane.
Convert g octane to mols octane. mol = g/molar mass.
Using the coefficients in the balanced equation, convert mols octane to mols CO2 and using PV = nRT and the conditions listed convert mols CO2 to L.
This should get you started. Post your work if you need further assistance.
If your car gets 21.4 mpg, how many L of CO2, will you release into the atmosphere, if you drove the 1280miles from Towson University to key West Florida for a much needed break after taking Chemistry 121? Assume 1.05atm and 17.0C.
What would your carbon load be in lbs (lbs of CO2 released) per mile traveled?
Assume that gasoline is octane; therefore the following reaction should be considered.
____C8H18(I) +_____O2(g)„³____CO2(g)+__H2O(g)
Physical Constants that you will find necessary:
Density of octane =0.7025g/ml
1Gal=3.79L
1LB=454g
9 answers
How to solve for gallons needed?
Can you please do the all things for me please, I really need help. Thank you so much.
21.4*x = 1280.
Solve for x.
Solve for x.
No. We don't do homework for you. We HELP you do your homework. You aren't asking for help; you're asking me to do your homework for you. We are tying to help so you will learn how to do these things yourself. If you need help you've come to the right place. If you need someone to do your work for you this is not the place.
gallons octane to L octane will be
#gallons x 3.785 liters right?
#gallons x 3.785 liters right?
The gallons I got 59.81
and the L octane 59.81*3.785= 226.38
and the L octane 59.81*3.785= 226.38
226.38 x 1000 = 226380mL
226380mL x 0.7025 g/mL = 159031.95grams.
I found those result is correct?
226380mL x 0.7025 g/mL = 159031.95grams.
I found those result is correct?
the answer is 9.2 L and 34 lbs.