How Do You Calculate the Mass of Tetraethyl Lead in Gasoline?

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    Chem Grade 11
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SUMMARY

The calculation of the mass of Tetraethyl Lead (Pb(C2H5)4) in gasoline involves determining the amount added per volume of gasoline and applying the density of Tetraethyl Lead. For 31.2 liters of gasoline, with an addition of 2.0 mL per 3.8 L, the mass can be calculated using the formula: Mass = Volume × Density. The density of Tetraethyl Lead is 1.653 g/mL, which is crucial for accurate mass determination. Additionally, to find the mass of elemental lead, the molar mass of lead must be utilized in conjunction with the mass of Tetraethyl Lead calculated.

PREREQUISITES
  • Understanding of basic chemistry concepts, including density and molar mass.
  • Familiarity with unit conversion, particularly between liters and milliliters.
  • Knowledge of how to calculate mass from volume and density.
  • Ability to perform calculations involving chemical formulas and stoichiometry.
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  • Learn how to calculate molar mass for complex compounds like Tetraethyl Lead.
  • Study the principles of density and its application in chemical calculations.
  • Explore unit conversion techniques between different measurement systems.
  • Investigate the environmental impact and regulations regarding the use of Tetraethyl Lead in gasoline.
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blazer105
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Hi I have two questions grade 11 chem questions that I don't know how to do. HELP PLEASE!

Q1: Tetraethyl( Pb(C2H5)4 ) lead was added to gasoline up to 2.0ml per 3.8L of gasoline.

a)Calculate the mass of Tetraethyl lead in the total amount of gasoline that you use for this trip (31.2 L of gasoline). The density of Tetraethyl lead is 1.653g/mL

b) Calculate the mass of elemental lead in the total amount of gasoline that you will consume for this trip.


I'm not sure but here is what I did.
Mass tetraethyl lead = 31.2 L gasoline * (2.0 mL Pb(C2H5)4/3.8 L gasoline) * 1.653 g Pb(C2H5)4
 
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Looks correct. Well, almost correct - not all units are OK. But that's a minor problem.

b is about using molar masses.
 
Yeah thanks for the help... I got the question like 12 mins after posting thanks for your help tho.
 

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