Question

Lets assume a car owner fuels his/her car with pure octane, how much CO2 (in mass / kg) goes into our atmosphere ? ( consider that an “average” person spends about $1000 per year on gasoline, and that gasoline currently costs on average $2.00 per gallon) Show your work.
1 gallon of gas weighs and average of 6 pounds.
1 kg is 2.2 pounds.

Answer 1

My teacher assigned us a different assignment but then changed it last minute and gave me an F for it. I have tried doing this, but I'm totally lost. Could someone help please?

Answer 2

@zepdrix
Can you help, no one is helping and I really would like some help...

Answer 3

Chemistry? :C
Ooo boy, I dunno... sorry

Answer 4

I don't know, but these people might be able to assist you: @JFraser @Vincent-Lyon.Fr @Photon336 @Rushwr @aaronq @JoannaBlackwelder @sweetburger

Answer 5

Also, welcome to OpenStudy! :)

Answer 6

Thanks!

Answer 7

Well, here's what I think. let's look @ this. we know the person spends 1,000 dollars a year on gas right? and that each gallon costs 2.00, so your first step should be figuring out how many gallons a person uses in a year do you know how to do this?

Answer 8

So 500 gallons a year

Answer 9

now. let's find the weight of our gas in kilograms. we need to convert gallons to pounds first. and then pounds to kilograms.

Answer 10

4 172.63 pounds so 1892.673131kg

Answer 11

or rounded, 1892.67kgs

Answer 12

now we know the weight of octane, we're asked how much CO2 is produced

Answer 13

How would we do this?

Answer 14

we need to go to the balanced equation for octane, reacting with oxygen. gasoline is being burned so this is combustion reaction.

1. now for this we need to convert kilograms of octane to moles of octane.

2. so you need to convert kilograms 1892.67 to grams first.

3. then convert grams of C8H8 to moles.

you do this by

4. grams* moles/(114.23 grams)

\[2 C_{8}H_{18} + 25 O_{2} = 18 H_{2}O + 16 CO_{2}\]

Answer 15

so you have 1892.67 kilograms.

1. this must be in grams
2. you must convert this number to moles of C8H18 by multiplying by the 1/114.23 molar mass of octane.

Answer 16

So in grams it would be 1892670
And if you have to muliplty this by 1/114.23 it would equal 16568.94
I'm not sure I did it correctly

Answer 17

these factors are all a set of conversion factors. set up the conversions with the units, and the math is very simple. it's all a set of fractions

Answer 18

so if I set the decimals it would be 16.5689?

Answer 19

you need to set this up so that you don't get confused this is dimensional analysis be sure to set it up like this because if you don't keep track of units you'll just get confused.

notice that the unit i want to find is in the numerator and the unit i dont find gets canceled out. when i multiply by the conversion factor.

\[\frac{ $1000 }{ $2.00 } = 500 gallons\]

\[500gallons*\frac{ 6pounds }{ gallon } = 3,000 pounds*(\frac{ 1kg }{ 2.2pounds}) =2,727 kilograms\]

\[2,727 kilograms*(\frac{ 1,000grams }{ kilograms })= 2,727,000 grams\]

this means we have 2,727,000 grams of Octane.

now let's convert this to moles of octane

\[2,727,000 grams*(\frac{ mole }{ 114.42grams}) = 23,833 mol Octane\]

Answer 20

the last thing we need to do is multiply by the molar ratio, I wont give that answer but I'll guide you in what to do.

\[moles of octane*(\frac{ 16CO_{2} }{ 2C_{8}H_{18} }) = Moles of CO2 \]

Answer 21

So far it makes sense, but how would we find the mass for this mole?

Answer 22

we need to take the number of moles of Octane found above and multiply it by the molar ratio. show us all how you do this. I gave you the guide above :)

Answer 23

23,833molOctane * (16CO2/2C8H18) = 190664molsof CO2

Answer 24

I attached the labels like you said to do, would this be correct?

Answer 25

yes. and you can find the number of grams or mass of CO2 by doing this


\[moles, of CO2* (\frac{ molar mass of CO2 }{ mol })\]

or in other words multiplying the moles you found by the molar mass of co2.

Answer 26

190664molsof CO2 * (190664molsof CO2 / 23,833molOctane) = 1525152g of CO2

Answer 27

\[190664 mol CO_{2}*(\frac{44.01grams }{ mol }) = 8,391,122 grams CO_{2}\]

Answer 28

What did I do wrong? Sorry I'm still learning...

Answer 29

make sure you multiply by the molar mass of CO2.

Answer 30

Oh ok! So then that's the mass right? Would it be better to convert it to kilograms?

Answer 31

yep can you show me how to do that i.e how you would set that up based on how I did it?

Answer 32

The whole thing? or just the part I messed up on?

Answer 33

To start you would find how many gallons a year this person needs.

\[$1000\div$2=500\]

Then you know that 1 gallon weighs 6 pounds so you multiply
\[500Gallons \times6Pounds =3000Pounds\]

Then to convert it to kgs you multiply[500Gallons \times6Pounds =3000Pounds \times(1kg)\div2.2lbs =2727kg\]

Then to convert it to grams to help find the mass you...
\[2727kg \times(1000g)=2727000g\]

Then we need to find molOctane so... (I couldn't figure out where to get the 144.42 from so I referred back to your sheets real quick)
\[27272000g \times114.42 =23833molOctane\]

Then to finally find the mass, you need to multiply by the molar mass of CO2 (16/2 were from the balanced equation)
\[23833molsOctane \times(16/2)=190664molsCO2\]

Then the final step is multiply with the molar mass...
\[190664molsCO2\times44.01molarmassofCO2=8391122gCO2\]

Then to make it easier, you convert to kg for a final answer of 8391.122kg of CO2 goes into the atmosphere per person per car.

Answer 34

I tried to do it myself without referencing your equations...

Answer 35

=D

Answer 36

Thank you so much for your help! I finally understand it a bit more! @Photon336

Answer 37

No problem