Based on my knowledge of stoichiometry, I will tell the driver that he or she has enough fuel for the race and as such does not need to do a pitstop. The amount of fuel needed to complete the race is about 1,690 gms of fuel. The car carries 3,500 gms of fuel.
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How will the knowledge of stoichiometry help in the calculations?</h3>
Recall that Density = mass / volume. This means that:
Mass = Density * Volume.
Therefore, the mass of 1 gallon of fuel = 700 g. We know this because:
Density of fuel = 700gms / 1 gallon.
Recall that the driver had 5.0 gallons in the tank. The mass of 5 gallons is:
5 × 700 = 3,500gms of fuel.
Please note that the rate at which the fuel is combusted is derived from the equation which has been given as:
C₅H₁₂ + 8 O₂ —> 6 H₂O + 5 CO₂
Please note that
1-mole of C₅H₁₂ needs 8 moles of O₂. If 1 mole of C₅H₁₂ has a mass = 72 gms, then
8 moles of O₂ has a mass = 256 gms. Following from the above, we can say that:
300 g of O₂ will require 300 × (72/256) g of C₅H₁₂ = 84.375 g of C₅H₁₂
That is 84.375gms of fuel will be burned or combusted by the car each lap;
To get 20 laps, we multiply the above figure by 20. That leaves us with
1, 687.5gms of fuel or approximately 1, 690 gms of fuel.
3,500 less 1,690 will leave us with an excess of 1,810 gms of fuel. Hence the driver has more than enough fuel to complete the race.
Learn more about stoichiometry at:
brainly.com/question/529998
