Question

Calculate the number of pounds of CO2 released into the atmosphere when a 12.0-gallon tank of gasoline is burned in an automobile engine. Assume that gasoline is primarily octane, C8H18 and that the density of gasoline is 0.692 g.mL-1 (this assumption ignores additives). Also assume complete combustion.

Useful conversion factors:
1 gallon = 3.785L
1 Kg = 2.204 lb

______________ lb

Answer 1

Answer:

106.95 lb of CO2

Explanation:

Equation of reaction:

2C8H18 + 25O2 = 8CO2 + 18H2O

Mass of C8H18 = density × volume = 0.692×12×3.785×1000 = 31430.64 g

From the equation of reaction above,

2 moles of C8H18 (228 g) produced 8 moles of CO2 (352 g)

31430.64 g of C8H18 will produce 31430.6×352/228 = 48524.50 g of CO2 = 48524.50/1000 = 48.52450 kg of CO2 = 48.52450×2.204 = 106.95 lb of CO2

Answer 2

Answer:

213.89 lb of CO2.

Explanation:

Equation for the reaction:

C8H18 + 25/2O2(g) --> 8CO2(g) + 9H2O(g)

Given:

Volume of gasoline = 12 gallon

Converting gallon to ml,

12 gallon * 3.785 l/1 gallon * 1000 ml/1 l

= 45420 ml

Density of the gasoline = 0.692 g/ml

Mass = density * volume

= 45420 * 0.692

= 31430 g

Molar mass of octane = (8*12) + (18*1)

= 114 g/mol.

Number of moles = mass/molar mass

= 31430/114

= 275.702 mol.

From the above equation, 1 mole of octane was completed burnt to give off 8 moles of CO2.

By stoichiometry,

Number of moles of CO2 = 275.702 * 8

= 2205.614 mol of CO2.

Molar mass of CO2 = 12 + (2*16)

= 44 g/mol

Mass of CO2 = number of moles * molar mass

= 2206.614 * 44

= 97047.02 g

Converting g to pound,

= 97047.02 g *1 kg/1000 g * 2.204 lb/1kg

= 213.89 lb of CO2.