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A compact car gets 37.5 miles per gallon on the highway. If gasoline contains 84.2 percent carbon by mass and has a density of 0.82059 g/mL, determine the mass of carbon dioxide produced during a 500 mile trip(3.785 litres per gallon)

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Answer:

127,849.7891 grams of of carbon dioxide produced during a 500 mile trip.

Step-by-step explanation:

Mileage of the compact car = 37.5 mile/gal

Miles covered through out the trip = 500 mile

Gallons of gasoline used up through out the trip =

=
(1)/(37.5)gal/mile* 500 mile=13.333 gallons=50.465 L

(1 gal = 3.785 L) given

Density of the gasoline =d= 0.82059 g/L

Mass of the gasoline = m

Volume of the gasoline ,V = 50.465 L = 50,465 mL


m= d* = 0.82059 g/L* 50,465 mL=41,411.074 g

Mass percentage of carbon = 84.2%

Mass of carbon in 41,411.074 g of gasoline be x.


84.2\%=(x)/(41,411.074 g)* 100

x = 34,868.1243 g


C+O_2\rightarrow CO_2

Moles of carbon =
(34,868.1243 g)/(12 g/mol)=2905.6770 mol

According to reaction , 1 mol carbon gives 1 mole of carbon dioxide.

Then 2905.6770 moles of carbon will produce:


(1)/(1)* 2905.6770 mol=2905.6770 mol of carbon dioxide

Mass of 2905.6770 moles of carbon dioxide:


2905.6770 mol* 44 g/mol=127,849.7891 g

127,849.7891 grams of of carbon dioxide produced during a 500 mile trip.

User Neil Walker
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