Asked by Coco

heating cyclopropane (C3H6) converts it to propane (CH2=CHCH3). The rate law is first order in cyclopopane. If the rate constant at a particular temp is 6.22x10^-4 s^1 and the concentration of cyclopropane is held at .0300 mol/L, what mass of propane is produced in 10.0 min in a volume of 2.50 Liters.
Can you please take me through this problem step by step, I have been trying different ways and now I am really confused.
Thanks C
Answered by DrBob222
Do you mean propEne? Do you mean k = 6.22E-4 s^-1?
rate = k*0.03 = mols/L*s
Then mols/L*s x (60 s/min) x 10 min x 2.5 L = mols in 2.5L in 10 min. Convert to grams. mols = grams/molar mass. You know molar mass and mols, solve for grams.
Answered by Coco
Ok I got .2799 moles, do i have to convert it to grams, how would I do that.
Answered by DrBob222
The problem asks for grams so I think you should make that conversion if you want to answer the question. For how you do it go back and read my response. I told you how to do it. Also you should punch those numbers into your calculator again. I didn't get that number. Also if your prof is picky about significant figures you will lose points because you have reported more than you are allowed.
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