Question

Calculate the number of pounds of CO2 released into the atmosphere when a 25.0-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.

Answer 1

Answer –

We are given, volume of gasoline = 25.0 gal

Density of gasoline = 0.692 g/mL

First we need to write the balanced combustion reaction of gasoline

2 C₈H₁₈ + 25 O₂ ------> 16 CO₂ +18 H₂O

Now we need to calculate mass of gasoline (C₈H₁₈)

We are given density in g/mL and volume in gal, so we need to convert volume gal to mL

We know,

1 gal = 3785.4 ml

So, 25 gal = ?

= 94635.3 mL

Now we know formula for calculating the mass from density and volume

Mass = density * volume

          = 0.692 g/mL * 94635.3 mL

         = 65487.6 g

Now we need to calculate moles of C₈H₁₈

Moles of C₈H₁₈ = mass of C₈H₁₈ / molar mass of C₈H₁₈

                          = 65487.6 g / 114.23 g.mol^-1

                          = 573.3 moles of C₈H₁₈

Now we need to calculate moles of CO₂ from the moles of C₈H₁₈

From balanced equation –

2 moles of C₈H₁₈= 16 moles of CO₂

So, 573.3 moles of C₈H₁₈ = ?

= 573.3 moles of C₈H₁₈ *16 moles of CO₂ / 2 moles of C₈H₁₈

= 4586.4 moles of CO₂

Now we need to calculate mass of CO₂

Mass of CO₂ = 4586.4 moles * 44 g.mol^-1

              = 2.018*10⁵ g

Now we need to convert g to lb

We know,

1 g = 0.002205 lb

So, 2.018*10⁵ g = ?

= 445 lb

445 pounds of CO2 released into the atmosphere when a 25.0-gallon tank of gasoline is burned in an automobile engine