How many moles, litres, grams of CO2 are emitted from gasoline

Answer 1

Gasoline is composed of a variety of alkanes, alkenes, and cycloalkanes, including one you may have heard of in your day to day life- octane. In order to answer the question, we'll need to put a few constraints on the problem and make a few assumptions.

Our assumptions

1. Gasoline is made up of 90% octane, by volume
2. The octane is burned with 100% efficiency
3. A tank of gasoline is 15 gallons and we will calculate the amount of CO2 released from burning a single tank

1) Converting gallons to liters

1 gallon = 3.79 liters (approx.)

15 gallons * (3.79 L/1gal) = 58.65 L of gasoline

2) Calculating volume of octane from volume of gasoline, assuming gasoline is 90% octane by volume

58.65 L gasoline * 0.90 = 51.71 L of octane

3) Calculating grams of octane (a quick search shows the density of octane is 0.703 g/mL)

51.71 L (1 mL/1*10-3 L) (0.703 g/ 1mL) = 35,970 g of octane

4) Calculating moles of octane

Octane is the IUPAC name for the chemical formula C8H18, therefore octane has a molar mass of 8*12.01 g/mol + 18*1.008 g/mol = 114.23 g/mol

35,970 g (1 mol / 114.23 g) = 314.9 mol

5) Converting moles of octane to moles of carbon dioxide

The reaction occurring to convert octane to carbon dioxide is called a combustion reaction and is characterized by reacting an alkane with oxygen gas to form carbon dioxide and water. The balanced equation for the combustion of octane is below.

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

314.9 mol of C8H18 (16 mol CO2 / 2 mol C8H18) = 2,519 mol of carbon dioxide

6) Calculating grams of carbon dioxide using the molar mass of CO2

2,519 mol (44.01 g/ 1 mol) = 110,900 grams of carbon dioxide

To conclude, if we assume a 15 gallon tank of 90% octane by volume gasoline is combusted completely, it would produce 110.9 kg of gaseous carbon dioxide.