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Homework Help: Finding the rate of production of O2 (g) from information given

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data

    2 KClO3 (s) + heat --> 2 KCl (s) + 3 O2 (g)

    If 0.20 g of KClO3 (s) decomposes in 2.8 s, at what rate is oxygen O2 (g) released over this time at SATP?

    3. The attempt at a solution

    0.20 g KClO3 x 1 mol/122.5495 g = 0.001632 mol KClO3

    Since 0.001632 mol decomposes in 2.8 seconds I found that 2 mol of it decomposes in 3431.37 seconds.

    So then to find the rate of decomposition of KClO3 I did 2/3431.37 = 0.000583 mol/s

    Now that I know the rate of decomposition of KClO3 I can find the rate of production of oxygen gas from the balanced reaction equation.

    0.000583 mol/s - 2 mol
    n - 3 mol

    n = 0.000874 mol/s O2 (g)

    The answer in the back says 22 mL/s but I have know idea how that answer was gotten. Can someone please help?
  2. jcsd
  3. Nov 17, 2013 #2
    Actually never mind I forgot to do one extra step with the PV = nRT, that was my error.
  4. Nov 17, 2013 #3
    Just a follow up question though, in a reaction such as 4 NO + 6 H2O -> 4 NH3 + 5 O2, where we know that the initial rate of NO is 0.050 mol/L*s so I tried to find the rate at which water is consumed and found it to be 0.075 mol/Ls. So if I have to explain why it is greater, I would start by saying that water is found in a greater number of moles, but we dont know the concentration though so how do know for sure that it has to have a greater rate?
  5. Nov 17, 2013 #4


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    Doesn't the ratio of consumption follow immediately from this:
    4 NO + 6 H2O​
  6. Nov 17, 2013 #5
    But how do we know how much concentrations we have of each?
  7. Nov 17, 2013 #6


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    Why does that matter? If one of the reagents is dilute, that will slow the reaction rate, but it won't alter the ratio between the quantities of the two agents consumed by the reaction.
  8. May 8, 2014 #7

    Can you please explain how you did this question?
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