You are a NASCAR pit crew member. Your employer is leading the race with 20 laps to go. He just finished a pit stop and has 5.0 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: The car uses an average of 300.0 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? (The density can be used as a conversion factor between grams and gallons)

Since there are 3500 g of fuel left in the tank, and he needs only 1687.5 g to complete 20 laps, he has enough fuel to complete the race. I will tell the driver that he does not need to make another pit stop as he has enough fuel to complete the race.

Explanation:

Density = mass / volume

Density of fuel = 700 g/ 1 gal

Therefore, the mass of fuel in 1 gallon = 700 g

The driver has 5.0 gallons of fuel in the tank.

The mass of 5.0 gallons of fuel = 5 × 700 = 3500 g of fuel

Equation of the combustion of fuel, C₅H₁₂ is given below:

C₅H₁₂ + 8 O₂ ---> 6 H₂O + 5 CO₂

1 mole C₅H₁₂ requires 8 moles of O₂

1 mole of C₅H₁₂ has a mass = 72 g

8 moles of O₂ has a mass = 256 g

Therefore, 300 g of O₂ will require 300 × (72/256) g of C₅H₁₂ = 84.375 g of C₅H₁₂

84.375 g of fuel is used by the car per lap;

20 laps will require 20 × 84.375 g of fuel = 1687.5 g of fuel.

Since there are 3500 g of fuel left in the tank, and he needs only 1687.5 g to complete 20 laps, he has enough fuel to complete the race. I will tell the driver that he does not need to make another pit stop as he has enough fuel to complete the race.

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