Organic Chemistry Roberts & Company Publishers presented by Sapling Learning Map

Question

Organic Chemistry Roberts & Company Publishers presented by Sapling Learning Map Calculate the number of pounds of CO2 released into the atmosphere when a 11.0-gallon tank of gasoline is burned in an automobile engine. Assume that gasoline is primarily octane, CeHle, and that the density of gasoline is 0.692 g mL-(this assumption ignores additives). Also assume complete combustion. Useful conversion factors: 1 gallon 3.785 L 1kg = 2.204 lb lb Exit O Previous Give Up & View Solution e Check Answer 0 Next > Hint

Explanation / Answer

volume of gasoline = 11.0 gallon
= 11.0*3.785 L
= 41.635 L
= 41635 mL

density = 0.692 g/mL

So,
mass of gasoline = density * volume
= 0.692 g/mL * 41635 mL
= 28811 g

Molar mass of C8H18,
MM = 8*MM(C) + 18*MM(H)
= 8*12.01 + 18*1.008
= 114.224 g/mol


mass(C8H18)= 28811 g

use:
number of mol of C8H18,
n = mass of C8H18/molar mass of C8H18
=(28811.0 g)/(114.224 g/mol)
= 2.522*10^2 mol
Balanced chemical equation is:
2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O


Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol

According to balanced equation
mol of CO2 formed = (16/2)* moles of C8H18
= (16/2)*252.2325
= 2017.8596 mol


use:
mass of CO2 = number of mol * molar mass
= 2.018*10^3*44.01
= 8.881*10^4 g
= 88.81 Kg
= 88.81*2.204 Kg
= 196 lb
Answer: 196 lb

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