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Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating enough energy to move a vehicle, or do other work.

How much CO2 and H2O (in grams) are produced in the combustion of 1.24 gallons of Octane? (density = 0.703 g/mL)

Helpful conversion, there are 4 quarts in a gallon. Beyond that, check out your equation sheet for other useful conversion factors!

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Answer:

  • 10.19 g CO₂
  • 4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

  • C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following conversion factors:

  • 1 gallon = 3.785 L
  • 1 L = 1000 mL

Now we convert 1.24 gallons to mL:

  • 1.24 gallon * [tex]\frac{3.785L}{1gallon} *\frac{1000mL}{1L} =[/tex] 4693.4 mL

We calculate the mass of Octane:

  • 4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the stoichiometric ratios and molecular weights to calculate the mass of CO₂ and H₂O:

  • CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * [tex]\frac{8molCO_{2}}{1molOctane}[/tex] * 44 g/mol =  10.19 g CO₂
  • H₂O ⇒ 3.30 g Octane ÷ 114g/mol * [tex]\frac{9molH_{2}O}{1molOctane}[/tex] * 18 g/mol = 4.69 g H₂O