Question

Driving 1000 miles a month is not unusual for a short-distance commuter. If your vehicle gets 19.70 mpg, you would use approximately 50.76 gallons of gasoline every month. If gasoline is approximated as C8H18 (density = 0.703 g/mL)...?

How many grams of carbon dioxide does your vehicle emit every month?

Answer 1

Each month, the car will produce 417 120 g of `CO_2`

Explanation:

First, let's get all the unit sorted out.

Assuming this is a US gallon, it will be equivalent to 3785 mL, and since the density of octane (representing the gasoline) is 0.703 g/mL, the one gallon would be

`3785 mLxx0.703g/(mL)=2661 g` per gallon

You have used 50.76 gallons each month, so

`50.76xx2661 = 135 072` grams each month.

The reaction that produces `CO_2` is

`2C_8H_18+25O_2rarr16CO_2+18H_2O`

so, we need to change those grams of gasoline into moles of octane:

Moles = `135 072g-:114g/"mol" = 1185 "mol"`

According to the reaction, we get 8 moles of `CO_2` for every 1 mole of `C_8H_18`, so the number of moles of `CO_2` produced will be

`1185xx8=9480` moles of `CO_2`

Since each mole of `CO_2` has a mass of 44.0 g, the total monthly output of `CO_2` will be...

`9480 "mol" xx 44.0 g/"mol"=417 120` g or 417.1 kg

Answer 2

About `4.171xx10^5` grams of `CO_2` is emitted.

Explanation:

[Step1] First, calculate the volume of gasoline `V` to drive 1000 miles.

`V= 50.76` (gallons)
Assuming gallon to be the US gallon( `1` US gal = `3.785` L),
`V=50.76 * 3.785 = 192.1` (L)

[Step2] Then, calculate the mass(g) and amount of substance(mol) of `C_3H_8` (octane).

The mass is `192.1 * 10^3` (mL) `* 0.703` (g/mL) `= 1.351xx10^5` (g).

Since the molar mass of `C_8H_18` (octane) is `12*8+1*18=114` (g/mol) , the amount of substance of octane is `(1.351xx10^5)/114= 1.185xx10^3` (mol)

[Step3] Calculate the `CO_2` emission.
The chemical equation for burning `C_8H_18` is
`2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O`. Eight moles of carbon dioxcide is emitted per one mole of octane.

Thus the amount of substance for `CO_2` is `1.185xx10^3*8=9.480*10^3`(mol).
The molar mass of `CO_2` is `12*1+16*2 =44`.
Therefore, the mass of `CO_2` emitted in a month is `9.480xx10^3*44 =4.171xx10^5` (g).