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2 moles of H20 heated at 10A for 1hr. What is the volume?

  1. Jun 18, 2015 #1
    1. The problem statement, all variables and given/known data
    2 moles of water undergoes electrolysis at 10A. What is the total volume of the system after 1hr?

    2. Relevant equations
    4 H+(aq) + 4e−→ 2H2(g)

    2 H2O(l) → O2(g) + 4 H+(aq) + 4e−

    Faraday's Law of Electrolysis
    Q = n(e-) x F

    F = 96500C

    3. The attempt at a solution

    Initial volume = 36mL

    Gas produced
    (10*3600s*4g)/(F*4)=0.373g H2

    (10*3600s*32g)/(F*4)=2.984g O2

    0.373g+2.984g=3.358g

    Final volume
    36mL+3.35mL =39.358mL
     
  2. jcsd
  3. Jun 18, 2015 #2

    Simon Bridge

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    3.358 rounds up to 3.36
    3.358g of gas does not make 3.358mL of gas.
    If you start with 36g of H2O and you remove some of the H and some of the O you do not end up with 36g.
     
  4. Jun 19, 2015 #3
    I apologize. I'm very new to chemistry. I thought I was solving for total volume including the expanding gas.

    Final volume
    36mL-3.35mL =32.65
     
  5. Jun 19, 2015 #4

    Borek

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    Staff: Mentor

    You can't calculate total volume of "the system" not knowing the pressure and the temperature. Perhaps it is just a matter of lousy wording of the question, or you are expected to assume STP.

    What you have found is the volume of water left. This is not total volume occupied.
     
  6. Jun 19, 2015 #5
    Yes, it was very poor wording. Still new to writing these equations. Thank you for your help.
     
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