I have more more problem i need some help figuring out. I really do not know how to even approach this problem; there is so much information.
- A newspaper article about the danger of global warming from the accumulation of greenhouse gases such as carbon dioxide states that "reducing driving your car by 20 miles a week would prevent the release of over 1000 pounds of C02 per year into the atmosphere". Is this a reasonable statement? Assume that gasoline is octane (C8H8) and that it is burned completely to CO2 and H20 in the engine of your car. The density of octane is .702/mL. There are 3.79L to one gallon, 2.20kg to one pound. Also assume that the average gas mileage in the United States is 22 miles per gallon. (Hint: you will start with one year and ehd with pounds of CO2)
Any ideas of the steps needed to complete this problem? It would be really appreciated.
Thanks
2C8H18 + 25O2 ==>16CO2 + 18H2O
20 mi/week x (52 weeks/1 yr)x (1 gal/22 mi) = ?? gallons/year.
Now convert gallons to liters and liters to kg, then go through the stoichiometry to how many kg of CO2 are released and change all that information back to pounds and compare with 1000 lbs.
I just spent an hour trying to find an error in some quick estimates I did on this BEFORE I FINALLY notice that your conversion factor for kg/lbs is not right. You should have 2.2 pounds = 1 kg and not the other way around. I am posting this hoping I can save you that hour of wasted time.
Opps. That must have been a mistake made by the teacher when typing because I just looked back at an extra i had of the worksheet and it is printed exactly as i typed it.
no worries. i did my best.
To determine whether the statement in the newspaper article is reasonable, we need to calculate the amount of CO2 released from driving 20 miles a week using the given information. We can break the problem down into several steps:
Step 1: Determine the volume of octane burned per week
To find the volume of octane burned, we need to calculate the number of gallons used when driving 20 miles. Given that the average gas mileage is 22 miles per gallon, we can use the following equation:
Gallons used per week = Miles driven per week / Gas mileage
Step 2: Convert the volume of octane to mass
Using the density of octane (.702 mL) and the conversion factor of 3.79 L per gallon, we can convert the volume of octane burned to mass:
Mass of octane used per week = Gallons used per week * Density of octane * Conversion factor
Step 3: Convert the mass of octane to CO2 produced
Given that octane (C8H18) is burned completely to CO2, we can use stoichiometry to determine the molar ratio between octane and CO2. Since octane has a molecular weight of 114.22 g/mol and CO2 has a molecular weight of 44.01 g/mol, the molar ratio is as follows:
Molar ratio = 8 moles of CO2 / 1 mole of octane
Using this molar ratio, we can convert the mass of octane used to the mass of CO2 produced:
Mass of CO2 produced per week = Mass of octane used per week * Molar ratio * Molecular weight of CO2
Step 4: Convert the mass of CO2 to pounds
Finally, to convert the mass of CO2 produced into pounds, we can use the conversion factor of 2.20 kg per pound:
Mass of CO2 produced per year = Mass of CO2 produced per week * Conversion factor
Now that we have the steps outlined, you can apply the given values and perform the calculations to determine the amount of CO2 released from driving 20 miles a week. Compare your answer to the figure mentioned in the newspaper article to determine if it is reasonable.