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
Can someone please please help me thanks.
Nancy, Jessica, anonymous?
See my response below under anonymous. You get better service here if you don't change screen names.
To calculate the amount of CO2 released into the atmosphere, we need to know the amount of gasoline consumed during the 1280-mile trip. We also need to consider the stoichiometry of the combustion reaction of octane to CO2 and H2O.
First, let's calculate the amount of gasoline consumed in liters. We can use the given fuel efficiency of 21.4 mpg:
Gasoline consumed (in gallons) = distance traveled (in miles) / fuel efficiency (in mpg)
Gasoline consumed (in gallons) = 1280 miles / 21.4 mpg = approximately 59.81 gallons
Next, we convert the gasoline consumed from gallons to liters:
Gasoline consumed (in liters) = gasoline consumed (in gallons) * 3.79 L/gallon
Gasoline consumed (in liters) = 59.81 gallons * 3.79 L/gallon = approximately 226.95 liters
Now, let's use the stoichiometry of the combustion reaction of octane to CO2 and H2O to determine the amount of CO2 produced.
From the balanced equation:
1 mole of octane reacts to produce 8 moles of CO2
To find the moles of octane:
Density of octane = 0.7025 g/mL
1 gallon = 3.79 L
Octane consumed (in grams) = gasoline consumed (in liters) * density of octane (in g/mL)
Octane consumed (in grams) = 226.95 L * 0.7025 g/mL = approximately 159.38 grams
To convert grams of octane to moles:
Molar mass of octane (C8H18) = 8 * atomic mass of carbon + 18 * atomic mass of hydrogen
Molar mass of octane = 8 * 12.01 g/mol + 18 * 1.01 g/mol = approximately 114.23 g/mol
Moles of octane = octane consumed (in grams) / molar mass of octane
Moles of octane = 159.38 g / 114.23 g/mol = approximately 1.395 mol
Since 1 mole of octane produces 8 moles of CO2, the moles of CO2 produced can be calculated as:
Moles of CO2 = 1.395 mol * 8 = approximately 11.16 mol
Lastly, we need to convert moles of CO2 to weight in pounds:
Weight of CO2 (in grams) = moles of CO2 * molar mass of CO2
Weight of CO2 (in grams) = 11.16 mol * (12.01 g/mol + 2 * 16.00 g/mol) = approximately 444.32 grams
Weight of CO2 (in pounds) = weight of CO2 (in grams) / 454 g/lb
Weight of CO2 (in pounds) = 444.32 g / 454 g/lb = approximately 0.98 lbs
Therefore, the carbon load in lbs per mile traveled would be:
Carbon load (in lbs/mile) = weight of CO2 (in pounds) / distance traveled (in miles)
Carbon load (in lbs/mile) = 0.98 lbs / 1280 miles = approximately 0.0008 lbs/mile
So, when driving from Towson University to Key West, Florida, your car would release approximately 0.0008 lbs of CO2 per mile traveled.