Question

1. A fuel has the formula C₉H₁₆ and its combustion is complete (i.e. it combines entirely with oxygen to produce CO₂ and water). Write down the balanced chemical equation of the combustion.

2. The fuel with formula C₉H₁₆ is used in a car that has a fuel efficiency of 20 mpg and is driven for 26,000 miles in one year. If one tree can assimilate 6 kg of CO₂ per year, would that tree offset the pollution caused by the car? Show all your calculations.

3. CO₂ and H₂O combine through the process of photosynthesis to form C₆H₁₂O₆ and O₂. Estimate the consumption of CO₂ (in kg per year) by a tree that produces 9 kg of O₂ per year.

4. Assuming that the world population is 7 billion and everyone in the planet plants one of the trees that produces 9 kg of O₂ per year; how much CO₂ would be capture in one year?

Answer 1

1.
When fuel is burnt it produces CO2 and H2O
C9H16 + O2   -----> CO2 + H2O
Now balance it,
C9H16 + 13O2   -----> 9CO2 + 8H2O

2.
Amount of fuel used for driving 26000 miles = 26000/20 = 1300 gallon
1 gallon = 3.7854 Kg
Fuel used = 1300*3.7854 Kg = 4921.02 Kg = 4921020 g
Molar mass of C9H16 = 124 g/mol
Number of moles of C9H16 = Mass / molar mass = 4921020 / 124 = 39685.6 mol

from reaction:
C9H16 + 13O2   -----> 9CO2 + 8H2O
1 mol of C9H16 gives 9 mol of CO2
So,
number of moles of CO2 produced = 9*39685.6 mol = 357170.8 mol
Molar mass of CO2 = 44 g/mol
So, mass of CO2 produced = number of moles * molar mass
     = 357170.8 * 44
    =1.57*10^7 g
    = 1.57*10^4 Kg
But tree can assimilate only 6 Kg in 1 year
so ,that tree will not offset the pollution caused by the car

3)
6CO2 + 6H2O ----> C6H12O6 + 6O2
mass of O2 produced = 9 Kg = 9000 g
Molar mass of O2 = 32 g/mol
number of moles of O2 = mass / molar mass
                                                = 9000/32
                                                = 281.25 mol
From equation above, 1 mol of O2 requires 1 mol of CO2
So, moles of CO2 consumed = 281.25 mol
Molar mass of CO2 = 44 g/mol
So, mass of CO2 consumed = number of moles * molar mass
     = 281.25 * 44
     = 12375 g
     = 12.375 Kg
Answer: 12.375 Kg/year

4)
AMount of Co2 captured in 1 year = 1 billion * 12.375 Kg
    =10^9 * 12.375 Kg
   = 1.2375*10^10 Kg