heating cyclopropane (C3H6) converts it to propene (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 propene is produced in 10.0 min in a volume of 2.50 Liters? I a, so lost on this. Any help is appreciated

Answer 1

ln(Ao/A) = akt
Ao = 0.03
a = 1
k = 6.22E-4
t = 10 min x (60s/min) - 600 s
Solve for (A) in mols/L
Plug in L to solve for mols
grams = mols x molar mass = ?

Answer 2

I got 517.5. I think I'm so off. Thanks anyway

Answer 3

Also im good at posting