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RE: #Chemistry Challenge 5

in #steemstem7 years ago (edited)
  1. C8H18 + 25/2 O2 --> 8 CO2 + 9H2O
  2. 45.6 euro could get 30.4 L isooctane
    implies 30.4 L could go 608 km
  3. Mass of isooctane = 22800 g
    implies mole of isooctane = 199.597 mole
    implies mole of CO2 = 1596.78 mole
    implies mass of CO2 emitted = 70.26 kg
    and since PV = nRT; Rearrange and get
    volume of CO2 emitted = 3.92 x 104 L
  4. 608 km takes 6.08 hr
    implies 5 L/hr
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1. Your equation is correctly balanced, but I would have preferred if you use the smallest possible whole numbers as factors in front of the element symbols. I count your answer as correct though.

2. Correct answer!

3. Why didn't you use the given (rounded) molar masses? In that case we would get 'nicer' numbers as results and I wouldn't need to verify every single of your calculations.

Your mass of isooctane is correct. Also n(C8H18) and n(CO2), but the mass of CO2 (even with not rounded molar masses) is a little bit too small. What did you use as M(CO2)? The volume (calculated with your numbers) is correct again (I guess you rounded R different than given).

4. The speed of chemical reactions isn't measured by volume / time!

I am just lazy so I use chemdraw to generate the molecular weigh LOL

More explanation on
Q3)
mole of CO2 = 1596.78 mole
implies mass of CO2 = 1596.78 * 44 = 70258 g = 70.26 kg
I use the given R btw
Q4
I thought the v(C8H18) where the v means volume LOL, but now i guess probably you mean speed
I would take rate of reaction as concentration over time usually, but I don't have concentration here, so I guess you are looking for change of mole per unit time?
If this is the case, here would be my answer:
5 L/ hr implies 32.89 mol/hr

Yes, right, but first you had another volume and also another mass of CO2. Better is to add a new comment if you change any parts of your solutions.

Concerning part 4: if any information seems to be missing you can use every given detail from parts 1 to 3 to find it out.
So far your answer is wrong.

How about:
v(C8H18) = - 32.89 mol/hr

NOT correct! :-)

But apart from that so far you did well ... :)

Edit: by the way, the reaction speed v is not negative (even if ∆c is negative), but that's not the point: your number is wrong, too.

humm... if you are expecting the answer in concentration:
[CO2] = 4.734 x 10-2 mol dm-3
implies rate of CO2 production = 6.7 x 10-3 mol dm-3 hr-1
implies speed of dissipating isooctane = 8.37 x 10-4 mol dm-3 hr-1

Thanks for letting us know about the concentration of carbon dioxide.

How exactly do you get
c(CO2) = 4.734 x 10-2 mol/dm3?

Nah.... sorry for getting so many typo
[CO2] = 1596.78 mol / 3.92 x 10 4 dm3
= 4.0734 x 10-2 mol dm-3
But the rest should be the same

m(CO2) is correct now, but please make a remark if you edit your text! (Same with your volume which was wrong at first.)
We need to know when you made your changes.

Sure Sure! Definitely !
Anyway, its seems not too many chemistry people around steemit, sadly, so it's really nice to meet you here :D
Btw, feel free to have a look my blog on some interesting chemistry sharing :D

:-)
I already follow you ...

Great! hope you enjoy it :D

Later I will check it more thoroughly!

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