)a vehicle owner fuels his/her car with pure octane. what quantity of co2 (in kg) goes into our atmosphere in one year of their automobile use? (this person spends $1500 per year on gasoline, and that gas costs $3.00 per gallon) show your work. 1 gallon of gas weighs 6 pounds. 1 kg is 2.2 pounds.

Money spent on fuel in 1 year is $1500, the cost of fuel is $3.00 per gallon. Thus, gallon of fuel used in 1 year can be calculated as follows:

$3 =1 gallon

$1=\frac{1}{3}gallons

Thus, $1500=\frac{1500}{3}gallons=500 gallons

converting this into m^{3}

1 gallon=0.0037 m^{3}

Thus,

500 gallons=1.8927 m^{3}

The fuel used is pure octane, the density of octane is 703 kg/m^{3} thus, mass of octane used can be calculated as follows:

m=d×V=703 kg/m^{3}×1.8927 m^{3}=1.33\times 10^{3} kg or 1.33\times 10^{6} g

Now, molar mass of octane is 114.23 g/mol

converting mass into number of moles as follows:

n=\frac{m}{M}=\frac{1.33\times 10^{6}g}{114.23 g/mol}=1.167\times 10^{4}mol

The reaction for combustion of octane is as follows:

2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2}+18H_{2}O

From the balanced reaction, 2 moles of octane gives 16 moles of CO_{2} thus, 1 mole of octane will give \frac{16}{2}=8 moles of CO_{2}.

1.167\times 10^{4}mol of octane =8\times 1.167\times 10^{4}mol=9.34\times 10^{4} mol of CO_{2}.

Now, molar mass of CO_{2} is 44.01 g/mol, mass can be calculated as follows:

m=n×M=9.34\times 10^{4} mol ×44.01 g/mol=4.11×10^{6} g.

converting into kg, mass of CO_{2} produced will be 4.11×10^{3} kg.

Thus, quantity of CO_{2} goes into our atmosphere in 1 year is 4.11×10^{3} kg.

the answers are

c and c