Calculating Limiting Reagent and Percent Yield for Gasoline Combustion

  • Thread starter hima
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In summary, the conversation is about solving a problem involving a chemical equation and various quantities of substances. The question involves determining the limiting reagent, calculating the mass of carbon dioxide produced, and finding the percent yield in a hypothetical experiment. The speaker also mentions the use of mole units and the need to convert from volume to mass.
  • #1
hima
1
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please i need to solve this problem, can you help me out. thank you.
equation: 2C8H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2O (l)
a)2.62kg(one gallon) of gasoline is burned with 7000.0L of oxygen gas at STP. what is the limiting reagent?
b) what mass of carbon dioxide is produced in part?
c) if 6.4 kg of carbon dioxide is collected in the lab using the amounts in part(a), what is the percent yield for this experiment?
 
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  • #2
Could you please tell us what you have done already, and where exactly you need help/don't understand?
 
  • #3
The questions are clear, but the problem is that we can't answer to homeworks here .
 
  • #4
You would perform most of your calculations in mole units. Since your given quantity of gasoline is in gallons(a volume unit), you need to know this gasoline's density (or specific gravity, at least something to work with); the density will give you the mass, and the mass will give you the moles (obviously using suitable conversion calculations).

The "l" in your reaction means "liquid", and the "g" in your reaction means "gas"
 

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