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Your car has 1.95 gallons of gasoline (octane, d = 0.6916 g/mL), which reacts with oxygen
according to the balanced reaction below. Your car uses the energy produced by this reaction
at a rate of 115 kJ per second while traveling at a speed of 65 miles per hour. Calculate the
distance (in miles) the car can travel using this amount of octane.
2 C8H18(g) + 25 O2(g) -----> 16 CO2(g) + 18 H2O(g) ΔHrxn = -10,900 kJ

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Answer:

38.3 miles

Explanation:

First, we convert 1.95 gallons to mililiters:

  • 1.95 gallons * [tex]\frac{3.785 L}{1gallon}*\frac{1000mL}{1 L}[/tex]= 7380.75 mL

Then we calculate how many grams of octane are available for the reaction, using its density:

  • 0.6916 g/mL * 7380.75 mL = 5104.53 g C₈H₁₈

Now we convert octane grams into octane moles, using its molar mass:

  • 5104.53 g C₈H₁₈ ÷ 114 g/mol = 44.78 mol C₈H₁₈

Then we calculate how many kJ are produced from the combustion of  44.78 mol C₈H₁₈, if 2 moles produce 10900 kJ:

  • 44.78 mol * 10900 kJ / 2 mol = 244032 kJ

We calculate how many seconds is the car available to keep going, if it spends 115 kJ per second:

  • 244032 kJ * 1 s / 115 kJ = 2122.02 s

We convert seconds to hours:

  • 2122.02 / 3600 = 0.59 hours

Finally we calculate the distance:

  • 65 mi/hour * 0.59 hour = 38.3 mi

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