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Homework Help: Chemistry redox reaction - gases mixing

  1. May 6, 2014 #1
    1. The problem statement, all variables and given/known data
    How many milliliters of Cl2 gas, measured at 28.0 °C and 750 torr, are needed to react with 18.5 mL of 0.173 M NaI if the I- is oxidized to IO3- and the Cl2 is reduced to Cl-?

    2. Relevant equations

    3. The attempt at a solution

    So the molar mass is .0030906 mol NaI. Now I need to find out how many moles of Cl2 are needed to react completely with that.
    So I need the balanced chemical equation.

    Cl2 + NaI --> IO3- + Cl-


    2e- + Cl2 --> 2Cl-

    NaI --> IO3- + Na+

    This is where I get stuck.
    I can't balance NaI as a redox reaction because I don't know the charges on that side. How do I balance the second part of the reaction?
  2. jcsd
  3. May 6, 2014 #2
    Is 3Cl2 + NaI + 3H2O --> IO3- + Na+ + 6H+ + 6Cl-

    the correct balanced reaction?
  4. May 6, 2014 #3


    User Avatar

    Staff: Mentor

    Technically - yes. But why don't you treat NaI as a dissociated salt (Na++I-), it is dissolved, isn't it?
  5. May 6, 2014 #4
    Ahh that's what was confusing me. I got the rest of it, thanks.
  6. May 6, 2014 #5

    what was the solution, im stumped
  7. May 7, 2014 #6
    So after you have the Redox reaction you have your ratio which is that for every 1 mol of NaI you have 3 moles of Cl2.

    You know that you have 18.5 mL of 0.173 M NaI so therefore

    .0185 L (.173 mol NaI/L) = moles of NaI.

    Now multiply that by three to get the moles of Cl2 because as shown from the balanced chemical equation - you need 3 moles of Cl2 for every 1 mol of NaI.

    From there you have


    Make sure to convert P to atm and T to Kelvin. n Is the amount of moles of Cl2 - solve for V. Done.

    Edit: oh yeah - convert to mL since that's what the question was asking for.
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