a. (3 pts) Your car should get 30 mi/gal running on octane (CsH18) and you are g

Question

a. (3 pts) Your car should get 30 mi/gal running on octane (CsH18) and you are going to drive 13 miles to your friends house. You particularly purchased this car because you are concerned about your carbon footprint- the mass of CO2 that comes out of your tailpipe. What should your carbon footprint be in terms of kilograms of CO2 to get to your friends house? Octane has a density of 0.703 g/ml. There are 3.785L in 1 gal. CsH18 (1) + 02 (g) ? CO2 (g) + H2O (g) (Unbalanced!) b. (2 pt) If your car actually produced 3.23 kg of CO2, what was your actual mileage (mile/ gallon?)

Explanation / Answer

a) The distance covered is 13 miles; the mileage of the car is 30 miles/gal; thus, 1 gal of octane (C8H18) can make the car run for 30 miles.

Therefore, gal(s) of C8H18 required to cover the 13 miles distance = (distance covered)/(mileage) = (13 miles)/(30 miles/gal) = 0.4333 gal (I shall keep a few guard digits extra).

The density of C8H18 is given as 0.703 g/mL. We know that

density = mass/volume

===> mass = volume*density.

Since we have 0.4333 gal of C8H18 (gallon is a unit of volume), we can easily calculate the mass of octane consumed. Therefore,

mass of C8H18 corresponding to 0.4333 gal octane = (0.4333 gal)*(0.703 g/mL) = (0.4333 gal)*(3.785 L/1 gal)*(1000 mL/1 gal)*(0.703 g/mL) = 1152.9485 g.

We shall need to find out the gram molar masses of C8H18 and CO2. The atomic masses are

C: 12.011 u

H: 1.008 u

O: 15.999 u

Gram molar mass of C8H18 = (8*12.011 + 18*1.008) g/mol = 114.232 g/mol.

Gram molar mass of CO2 = (1*12.011 + 2*15.999) g/mol = 44.009 g/mol.

Mole(s) of C8H18 corresponding to 1152.9485 g = (mass of C8H18)/(molar mass of C8H18) = (1152.9485 g)/(114.232 g/mol) = 10.0930 mole.

Balance the given reaction as below.

2 C8H18 (l) + 25 O2 (g) ----------> 16 CO2 (g) + 18 H2O (g)

As per the stoichiometric equation,

2 moles C8H18 = 16 moles CO2.

Therefore, 10.0930 moles C8H18 = (10.0930 mole C8H18)*(16 moles CO2/2 moles C8H18) = 80.744 moles CO2.

Carbon footprint in terms of kilograms of CO2 produced = (80.744 moles)*(44.009 g/mole)*(1 kg/1000 g) = 3.5535 kg ? 3.55 kg (ans).

Note that 1 L = 1000 mL and 1 kg = 1000 g.

b) However, in reality, the carbon footprint was 3.23 kg; therefore, moles of CO2 corresponding to 3.23 kg = (mass of CO2 produced in grams)/(molar mass of CO2) = (3.23 kg)*(1000 g/1 kg)/(44.009 g/mol) = 73.3941 mole (guard digits extra).

Again, as per the stoichiometric equation,

16 moles CO2 = 2 moles C8H18.

Therefore, 73.3941 moles CO2 = (73.3941 moles CO2)*(2 moles C8H18/16 moles CO2) = 9.1743 moles C8H18.

Mass of C8H18 corresponding to 9.1743 moles in grams = (9.1743 mole)*(114.232 g/mol) = 1044.9144 g.

Volume of C8H18 in gal corresponding to 1044.9144 g C8H18 = mass/density = (1044.9144 g)/(0.703 g/mL) = 1486.3647 mL = (1486.3467 mL)*(1 L/1000 mL)*(1 gal/3.785 L) = 0.3927 gal (guard digits extra).

Therefore, the mileage of the car = (distance covered)/(gals of octane required) = (13 miles)/(0.3297 gal) = 39.4298 miles/gal ? 39.4 miles/gal (ans).

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