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
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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.
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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
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Thanks for letting us know about the concentration of carbon dioxide.
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How exactly do you get
c(CO2) = 4.734 x 10-2 mol/dm3?
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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
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:)
This result is the (lower) concentration of CO2 when it was emitted as a gas already ... but gasoline is a liquid ...
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Ya, so why I am wondering what unit you are expecting
Speed of reaction: defined as how quickly or slowly a reaction takes place
(So my very initial attempt was in liter per hour)
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As you wrote yourself, reaction speed is measured in v = ∆c/∆t (and if it is an educt: v = - ∆c/∆t, because ∆c is negative, so with the additional minus v itself stays positive).
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