My motorcycle has a capacity of 4.5 gal of gasoline. If I combust all of the octane to power my motorcycle, what is the volume of CO2 generated from just my motorcycle on one tank of gas?Assume that I burn all of the gasoline into CO2 on warm day in June, where it is 85 degrees F outside with a barometric pressure of 30 mmHg
How much of this do you know how to do? Show what you can do and explain where you need help.
thanks so much for your help Dr. Bob I need to know how to set up the problem. I converted gallons to liters and temperature to kelvin and the inHg to atm. Other than that, I don't know how to set up this problem
I do have the added note that the answer to this problem is not the volume of the atmosphere - instead you are determining the volume of the atmsophere displaced by the CO2 generated from my motorcycle exhaust
1. Write and balance the equation for the combustion of octane. I will do that to get you started.
2C8H18 + 25O2 ==> 16CO2 + 18H2O
2. Look up the density and use that to convert liters gasoline to grams.
3. Convert grams octane to mols. mols = grams/molar mass.
4. Using the coefficients in the balanced equation, convert mols octane to mols CO2
5. Use PV = nRT to convert mols CO2 at the conditions listed to volume in liters. If you want the answer in liters you are done.
that's all I needed thank you so much, turns out I was way off
I'm not sure I understand the "added note". What I have given you at the end is liters CO2 generated by 4.5 gallons of gasoline at 85 F and 30 in Hg. Of course it will displace the atmosphere.
To calculate the volume of CO2 generated from burning a given amount of gasoline, you need to know the chemical formula of gasoline, the combustion reaction, and the stoichiometry of the reaction. Here's how you can do it step by step:
1. Convert the temperature from Fahrenheit to Kelvin:
T(K) = (T(°F) + 459.67) × 5/9
T(K) = (85 + 459.67) × 5/9
= 302.038 K
2. Convert the barometric pressure from mmHg to atm:
P(atm) = P(mmHg) / 760
P(atm) = 30 mmHg / 760
= 0.0395 atm
3. Calculate the number of moles of gasoline using the ideal gas law:
PV = nRT
V = 4.5 gal = 17.034 L (assuming 1 gal = 3.78541 L)
R = 0.0821 L·atm/(K·mol) (ideal gas constant)
n = (P(atm) × V) / (R × T(K))
n = (0.0395 atm × 17.034 L) / (0.0821 L·atm/(K·mol) × 302.038 K)
≈ 0.875 moles
4. Determine the balanced chemical equation for the combustion of gasoline. The combustion reaction for octane (a component of gasoline) is:
2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O
This indicates that 2 moles of octane (C8H18) produce 16 moles of CO2.
5. Calculate the number of moles of CO2 produced using the stoichiometry of the reaction:
Moles of CO2 = (n of gasoline) × (moles of CO2 / moles of octane)
= 0.875 moles × (16 moles CO2 / 2 moles octane)
= 7 moles CO2
6. Convert moles of CO2 to volume at standard temperature and pressure (STP). STP is defined as 1 atm pressure and 273.15 K temperature.
1 mole of any ideal gas at STP occupies 22.4 L.
Volume of CO2 at STP = (moles of CO2) × (22.4 L/mole)
= 7 moles × 22.4 L/mole
= 156.8 L
Therefore, by combusting one tank of gasoline in your motorcycle, you would generate approximately 156.8 liters of CO2 under the given conditions.