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Homework Help: Weird Stoichiometry problem

  1. Nov 23, 2014 #1
    Hello, Im in desperate need of help with setting up this problem and I would very much appreciate being
    steered in the right direction.

    1. The problem statement, all variables and given/known data

    Find the mass percent of Fe3+ in a 0.0293g sample of iron ore if 13.25 ml of a .200 M stannous chloride solution is required to react completely with Fe3+.

    2. Relevant equations
    2 Fe3+(aq) + Sn2+(aq) ⇒2Fe2+(aq)+Sn4+

    is the net ionic equation provided

    3. The attempt at a solution
    So I am trying to find M of Fe3+ so

    g Fe3+→mol Fe/g Fe→mol Fe/mol SnCl2→.200molSnCl2/1L ......etc

    Sorry I dont really know how to represent a dimensional analysis problem. Basically I am trying to end up with mol/L to calculate the molarity for Fe3+ but I have a strong suspicion that I am attempting this problem in the wrong way. I have tried to rework this problem several times sometimes starting with the 13.25ml but I am ending up at a dead end every time.

    Any help or advice is sincerely appreciated.
     
  2. jcsd
  3. Nov 23, 2014 #2

    Bystander

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    One step at a time. You have 13.25 ml of what, and what information does that give you?
     
  4. Nov 23, 2014 #3
    Its 13.25 ml of stannous chloride solution which contains 0.200 moles per liter. It is the amount of solution that is required to completely react with 0.0293g Fe3+.
     
  5. Nov 23, 2014 #4

    Bystander

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    Not two steps at a time --- ONE.
    What information?
     
  6. Nov 23, 2014 #5
    How many moles of stannous ions are present in 13.25 ml of 0.200 molar stannous chloride?

    Chet
     
  7. Nov 24, 2014 #6
    Thanks for the response. It gives me the information about the number of moles contained in 13.25 ml solution.

    2.65x10-3 moles. Ive tried to use this: 13.25ml x 1L/103ml x.200mole SnCl2/ 1 L

    but I keep having trouble ending up with mol/L to find the molarity of Fe3+. Ill keep trying to play with the equation to see what I get. Thank you.
     
    Last edited: Nov 24, 2014
  8. Nov 24, 2014 #7

    Bystander

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    Moles of what?
     
  9. Nov 24, 2014 #8
    Chet[/QUOTE]
    Oh sorry, moles of stannous chloride.
     
  10. Nov 24, 2014 #9

    Bystander

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    So, now you know how many moles of stannous chloride reacted with Fe3+, and you've been given a reaction that tells you the proportional relationship between Sn2+ and Fe3+. Step 2 is using the number you just found with that relationship.
     
  11. Nov 24, 2014 #10
    Yes the problem I am having is utilizing this relationship in dimensional analysis to end with the units I am looking for. Here is one of the strategies I have come up with (the main problem I am having is fitting in the 0.0293g Fe3+ )

    2.65x10-3 moles SnCl2 x 2 mol Fe3+/1mol SnCl2 x 55.85g Fe3+/1 mol Fe3+

    I keep trying to cancel out units to end up with mol Fe3+/L but I feel as though I am approaching the dimensional analysis set up in the wrong way.
     
  12. Nov 24, 2014 #11

    Bystander

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    What is the point of looking for mol Fe/L

    when this is the question you're trying to answer?

    You have yet to write an equation.
     
  13. Nov 24, 2014 #12
    Well mol/L is the mass percent of a solution right? I am guessing this is wrong.

    Yeah I tried to use this to represent a dimensional analysis chain : 2.65x10-3 moles SnCl2 x 2 mol Fe3+/1mol SnCl2 x 55.85g Fe3+/1 mol Fe3+.......
     
  14. Nov 24, 2014 #13

    Bystander

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    You are interested in solving a pair of proportions, "a is to b as c is to d;" you've seen such proportions written as, " a:b::c:d," or as an equation (includes an equal sign) "a/b = c/d." For the first proportion, you have been given "a" (moles of Fe), "b" moles of Sn, "d" moles of Sn to titrate an unknown quantity of Fe. For the second, you are calculating "a" (mass of Fe3+ from the first proportion), given "b" (mass of sample), and calculating a numerical value for "c" (number of %), and know "d" (100).

    Do these proportions ONE step at a time, and once you understand the steps, you can solve the proportions for individual values of a, b, c, or d as necessary and substitute them into previous or subsequent proportions to generate a lumped dimensional equation.
     
  15. Nov 24, 2014 #14
    Im sorry Im such an idiot for not understanding...but I just dont understand this problem, Im even more confused now. Thank you so much for your patience.
     
  16. Nov 24, 2014 #15
    You are not trying to find the amount of Fe3+ is any solution. You are trying to find it in the solid ore, that is, the mass percent in the solid ore. You have the right idea in trying to calculate the number of grams of Fe3+, but you made a mistake when you said there are 2 moles of Fe3+ for every mole of stannous. There is only 1/2 mole of Fe3+ for every mole of stannous.

    Chet
     
    Last edited by a moderator: Nov 24, 2014
  17. Nov 24, 2014 #16

    Borek

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    No, it is a concentration, but not mass percent.

    Besides, you don't need concentration - you need amount of Fe. That can be easily calculated from teh stoichiometry of the reaction and known amount of tin reaction.

    I have a feeling you are paralyzed by trying to blindly apply dimensional analysis and without understanding what is happening first. Bystander suggestion of doing calculations a step at a time is perfect.
     
  18. Nov 24, 2014 #17
    Why are there 1/2 mole Fe for every 1 mole of stannous if the balanced equation is : 2 Fe3+(aq) + Sn2+(aq) ⇒2Fe2+(aq)+Sn4+ ?


    So I am trying to find grams of Fe? And yes I I am blindly trying to apply dimensional analysis to this problem because I dont understand what to do. There are no example problems like this in the book. When I created the thread I thought I was heading in the right direction but now I dont even know.
     
  19. Nov 24, 2014 #18

    Bystander

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    Okay, let's clarify some background for purposes of us (Bystander, Chet, Borek) helping you. What's the highest level of math you've completed?
     
  20. Nov 24, 2014 #19
    Algebra 2 :/
     
  21. Nov 24, 2014 #20

    Bystander

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    Thank you --- give me a couple minutes to double check myself regarding what all I'm taking for granted and rethink how to talk you through this. "I shall return."
     
  22. Nov 24, 2014 #21
    Oops. You're right. I must have misread the equation.

    Yes, and you've already shown how to do that. So, what number do you get? What fraction is this of the mass of the ore? What percent?

    Chet
     
  23. Nov 24, 2014 #22
    Okay so in 13.25 ml of a .200 M stannous there are 2.65x10-3 moles. So I convert moles SnCl to moles Fe to grams Fe?

    2.65x10-3moles SnCl2 x 2 moles Fe/1 mol SnCl2 x 55.85g Fe/1 mole Fe to end up with 0.296 grams Fe? So this is 29.6% of the mass of Fe? Is this what I am looking for?
     
    Last edited: Nov 24, 2014
  24. Nov 24, 2014 #23

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    Good to here. Now, what's the mass of the sample?
     
  25. Nov 24, 2014 #24
    the originally mass of the sample was 0.0293g Fe. So am I supposed to divide the actual by the theoretical yield and multiply by 100% to get the mass percentage?
     
  26. Nov 24, 2014 #25

    Bystander

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    ... and, who wrote this problem? Please tell me you transcribed 0.0293 g from 0.0293 kg. And why didn't I work it through first and catch this disaster?

    "Houston, we have a problem." If there's a larger mass of ferric iron in the analysis than there was in the original sample, it's about par for textbook writing and proofreading these days.

    Please double check the units in the original problem statement for us. One of us shoulda caught that earlier --- me particularly.
     
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