A website promoting the use of alternative energy vehicles and hybrid technologies claims that, \\"A typical...

Free

80.2K

Verified Solution

Question

Chemistry

A website promoting the use of alternative energy vehicles andhybrid technologies claims that, \\"A typical automobile in the USAuses about 500 gallons of gasoline every year, producing about 5tons of carbon dioxide.\\" To determine the truth of this statement,calculate how many tons of carbon dioxide are produced when 500.0gallons of gasoline are combusted. Assume that the primaryingredient in gasoline is octane, C8H18(l), which has a density of0.703 gÂ·mL^ â€“1
HINT:
Generate a balanced chemical equation for the combustion ofoctane. The reactants will be C8H18+ O2, and the products will beCO2+ H2O. Using the mole ratios from the balanced equation, performthe correct unit conversions to arrive in tons of carbon dioxide.Note that : 1 gallon = 3785.4 cm3 1 lb = 453.6 g

Answer & Explanation Solved by verified expert

4.0 Ratings (465 Votes)

1 USA gallon = 3.7854liters

Â Â Â Â Â Â Â 500gallon = 500*3.7854 = 1892.7liters

mass of octane = volume * density

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1892.7*703 = 1330568g of octane
molar mass of octane = 114g/mole

no of moles of octane = W/G.M.Wt = 1330568/114 = 11671.65 moles
2C8H18+ 25O2--------> 16Co2 + 18H2O
2 moles of C8H18 on combustion to produced 16 moles of Co2

11671.65 moles of C8H18 to gives = 16*11671.65/2 =93373.2moles of Co2
mass of Co2 = no of moles * gram molar mass

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 93373.2*44 = 4108420.8g = 4108KgÂ Â = 4108*0.00110231 = 4.53tons
Â Â Â Â Â Â Â Â Â Â Â Â Â

Get Answers to Unlimited Questions

Join us to gain access to millions of questions and expert answers. Enjoy exclusive benefits tailored just for you!

Membership Benefits:

Unlimited Question Access with detailed Answers
Zin AI - 3 Million Words
10 Dall-E 3 Images
20 Plot Generations
Conversation with Dialogue Memory
No Ads, Ever!
Access to Our Best AI Platform: Flex AI - Your personal assistant for all your inquiries!

Become a Member