stoichiometry solution

Question

estimate the amount of CO2 (in kg ) that you emit each year by driving. estimates the amount of CO2 (in kg ) that you emit each year by breathing. Come up with another question that you can use stoichiometry to answer, and answer it.show work

Answer 1

These are estimation problems, so the marks are in the method and in stating your assumptions, not in matching some exact number. State every assumption explicitly, then let stoichiometry do the rest. Substitute your own mileage and diet and your answer will differ - that is fine and expected.

Part 1 - CO2 from driving

Assumptions: 12,000 miles per year, 25 miles per gallon, gasoline treated as octane C8H18 with density 0.75 kg/L.

Fuel volume: 12,000 mi / 25 mpg = 480 gal, and 480 gal x 3.785 L/gal = 1817 L

Fuel mass: 1817 L x 0.75 kg/L = 1363 kg = 1.363 x 10^6 g

Moles of octane: 1.363 x 10^6 g / 114.23 g/mol = 11,930 mol

Balanced combustion: 2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O, so 8 mol CO2 per mol octane.

Moles CO2: 8 x 11,930 = 95,440 mol

Mass CO2: 95,440 mol x 44.01 g/mol = 4.20 x 10^6 g = 4200 kg per year

Sanity check: that is about 3.1 kg of CO2 per kg of fuel burned, and roughly 4.2 metric tons a year. The EPA figure for an average passenger car is about 4.6 metric tons, so the estimate lands in the right place. Notice the CO2 outweighs the fuel by a factor of three, because every carbon atom picks up two oxygen atoms from the air.

Part 2 - CO2 from breathing

Assumptions: a 2000 kcal per day diet, metabolised as glucose.

C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O, releasing 2803 kJ per mole

Energy: 2000 kcal x 4.184 kJ/kcal = 8368 kJ per day

Moles glucose: 8368 kJ / 2803 kJ per mol = 2.99 mol per day

Moles CO2: 6 x 2.99 = 17.9 mol per day

Mass: 17.9 mol x 44.01 g/mol = 788 g per day

Per year: 788 g x 365 = 2.88 x 10^5 g = 288 kg per year

Independent check, via breath volume: about 0.5 L per breath at 15 breaths per minute is 10,800 L of exhaled air per day. Exhaled air is roughly 4% CO2, giving 432 L of CO2, and at 24.5 L/mol that is 17.6 mol per day, or 776 g. The two completely different routes agree to within 2%, which is far better evidence that the method is sound than either number alone.

Worth noting: breathing releases only about 7% as much CO2 as driving does, and unlike the car it is carbon-neutral - it came from food grown by plants that took the same carbon out of the air last season.

Part 3 - my own question: how much CO2 comes from the cement in a concrete driveway?

I like this one because most of the CO2 is not from burning anything - it comes straight out of a decomposition reaction, which makes it a pure stoichiometry problem.

Assumptions: a driveway 6 m x 3 m x 0.10 m; concrete density 2400 kg/m^3; cement is 12% of concrete by mass; clinker is 65% CaO by mass.

Volume: 6 x 3 x 0.10 = 1.8 m^3, so mass = 1.8 x 2400 = 4320 kg concrete

Cement: 0.12 x 4320 = 518 kg, of which CaO is 0.65 x 518 = 337 kg

The kiln reaction is calcination: CaCO3 -> CaO + CO2, one mole of CO2 per mole of CaO.

Moles CaO: 3.37 x 10^5 g / 56.08 g/mol = 6010 mol

Mass CO2: 6010 mol x 44.01 g/mol = 2.6 x 10^5 g = 260 kg

Add roughly the same again for the fuel burned to heat the kiln to 1450 degrees C, and the real figure is about 500 kg.

Why the answer is interesting: that driveway carries about six weeks of your driving emissions, and half of it can never be engineered away by switching fuels, because it is the chemistry of the material itself. Cement production is around 8% of global CO2 emissions for exactly this reason. A single balanced equation, CaCO3 to CaO plus CO2, explains a problem that no amount of efficiency improvement can solve.