Answer:
127,849.7891 grams of of carbon dioxide produced during a 500 mile trip.
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 =
= [tex]\frac{1}{37.5}gal/mile\times 500 mile=13.333 gallons=50.465 L[/tex]
(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
[tex]m= d\times = 0.82059 g/L\times 50,465 mL=41,411.074 g[/tex]
Mass percentage of carbon = 84.2%
Mass of carbon in 41,411.074 g of gasoline be x.
[tex]84.2\%=\frac{x}{41,411.074 g}\times 100[/tex]
x = 34,868.1243 g
[tex]C+O_2\rightarrow CO_2[/tex]
Moles of carbon = [tex]\frac{34,868.1243 g}{12 g/mol}=2905.6770 mol[/tex]
According to reaction , 1 mol carbon gives 1 mole of carbon dioxide.
Then 2905.6770 moles of carbon will produce:
[tex]\frac{1}{1}\times 2905.6770 mol=2905.6770 mol[/tex] of carbon dioxide
Mass of 2905.6770 moles of carbon dioxide:
[tex]2905.6770 mol\times 44 g/mol=127,849.7891 g[/tex]
127,849.7891 grams of of carbon dioxide produced during a 500 mile trip.
