Question

You are a NASCAR pit crew member. Your employer is leading the race with 15 laps to go. He just finished a pit stop and has 3.0 gallons of fuel in the tank. On the way out of the pits, he asks, “Am I going to have enough fuel to finish the race or am I going to have to make another pit stop?” You whip out your calculator and begin your calculations based on your knowledge of stoichiometry. Other information you know is:

C5H12 + O2 → CO2 + H2O

The car uses an average of 275.0 grams of O2 for each lap.
The formula for fuel is C5H12
The fuel has a density of 700 g/gal.

What do you tell the driver? Can he finish the race? Will he have fuel left over?

Answer 1

When Darrell is radioed back he would be asked to “Go for it!”

Step-by-step explanation:

Here we are given the fuel as C₅H₁₂, therefore the combustion reaction is given as

C₅H₁₂ + 8O₂ → 5CO₂ + H₂O

Mass of oxygen consumed on each lap = 300 g

Molar mass of oxygen gas O₂ = 32 g/mol

Number of moles of oxygen n 300 g of O₂ =

(300 g)/(32 g/mol) = 9.375 moles

For complete combustion, one mole of oxygen gas reacts with one mole C₅H₁₂, to form 5 moles of CO₂ and one mole of H₂O

Therefore 9.375 moles of oxygen ill combine with 9.375/8 or 1.172 moles of C₅H₁₂

Mass of 1.172 moles is given as

Mass = Number of moles × Molar mass

= 1.172 moles × 72.15 g/mole = 84.551 g

Therefore the mass of fuel to complete one lap = 84.551 g

However there are 25 gallons or 3.5 kg in the tank therefore we have

Number of laps the fuel in the tank can last = Mass of fuel in the tank/ Mass of fuel consumed per lap

= 3.5/84.551 = 41.395 laps.

Number of laps the fuel in the tank can last = 41.395 laps.

Since there are 20 laps left to complete, which is less than 41.395 laps left in the fuel tank of the vehicle, then Darrell would be asked to go for it.