# 2 C8H18(l) + 25 O2(g) -> 16 CO2(g) + 18 H2O(g) If you burned one gallon of gas (C8H18) (approximately 4000 grams), how many liters of carbon dioxide would be produced at a temperature of 21.0°C and a pressure of 1.00 atm?

Feb 1, 2015

7000 L.

#### Explanation:

One gallon of octane produces approximately 7000 L of carbon dioxide.

Note: At $\text{20"^("o")"C}$, the density of octane, ${\text{C"_8"H}}_{18}$, is 0.70300 g/mL, and the mass in grams in one gallon would be 2661 g. http://blueskymodel.org/gallon-gas At $\text{15"^("o")"C}$, the density of octane is 0.91786 g/mL (http://www.simetric.co.uk/si_liquids.htm), and the mass of one gallon of octane would be 3474 g, which is approximately 4000 g.

Since the temperature in the problem is $\text{21"^("o")"C}$, I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.

Step1. You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. Step 2. Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on the mole ratio between octane and carbon dioxide in the balanced equation. Step 3. Then use the ideal gas law to determine the volume in liters of carbon dioxide that can be formed.

Known/Given:
mass of octane = approximately 4000 g
temperature: $\text{21.0"^("o")"C}$ + $\text{273.15}$ = $\text{294.2 K}$
pressure = 1.00 atm
gas constant $\text{R}$ = ${\text{0.08205736 L atm K"^(-1)"mol}}^{- 1}$
molar mass of octane, ${\text{C"_8"H}}_{18}$ = $\text{114.232 g/mol}$
mole/mole ratio for ${\text{C"_8"H}}_{18}$ and $\text{CO"_2}$ = ${\text{2 mol" "C"_8"H}}_{18}$/$\text{16 mol" "CO"_2}$

Unknown:
volume of $\text{CO"_2}$

Balanced Chemical Equation:
$\text{2 C"_8"H"_18("l")}$ + $\text{25 O"_2("g")}$ $\rightarrow$ "16 O"_2("g") + "18 H"_2"O"("g")

Ideal Gas Law:
$\text{PV}$ = $\text{nRT}$

Step 1. Determine Moles of Octane in One Gallon
$\text{4000 g C"_8"H"_18}$ x $\text{1 mol"/"114.232 g}$ = $\text{35.0164 mol C"_8"H"_18}$

Step 2. Moles of Carbon Dioxide Produced by One Gallon of Octane
Multiply moles of octane times the mole/mole ratio between octane and carbon dioxide, so that carbon dioxide is in the numerator.

${\text{35.0164 mol C"_8"H}}_{18}$ x ($\text{16 mol" "CO"_2}$/${\text{2 mol" "C"_8"H}}_{18}$) = $\text{280.1312 mol" "CO"_2}$

Step 3. Calculate Volume of ${\text{CO}}_{2}$ Produced from One Gallon of Octane using the Ideal Gas Law, PV = nRT.
P = 1.00 atm
n = 280.1312 mol $\text{CO"_2}$
R= ${\text{0.08205736 L atm K"^(-1)"mol}}^{- 1}$
T = $\text{294.2 K}$

$\text{V}$ = $\text{nRT"/"P}$ = $\text{(280.1312 x 0.08205736 x 294.2)"/"1.00 atm}$ = $\text{6763 L}$ = $\text{7000 L}$ due to only one significant figure in 4000 g.