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Stoichiometry: Identify the metal and formula of it

  1. Aug 21, 2016 #1
    1. The problem statement, all variables and given/known data
    All of the lanthanide metals (La through Lu) react with HCL to form compounds having eithe the formula MCl2, MCl3, MCl4, (where M represents the metallic element). Each metal forms a single compound. A chemist has a 0.250g sample of a lanthanide metal, and she wishes to identify the metal. She reacts the metal with excess HCl and obtains 0.427g of the product. Based on this information, identify the metal and write the chemical formula of the product.

    2. Relevant equations


    3. The attempt at a solution
    I'm not sure the below equation is correct. I originally wrote the equation without the H2, but it did not seem to be right with conservation of mass.
    M + HCL —> MCl2 (or MCl3, MCl4) + H2
    Basically, if the H2 does belong there, then I'd just have to guess MCl2, 3, or 4, then balance the equation, then randomly guess an element right? Seems like a lot of work.
     
  2. jcsd
  3. Aug 21, 2016 #2

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  4. Aug 21, 2016 #3
    I can't see anything on your post
     
  5. Aug 21, 2016 #4

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    MCln is all you need; ignore the H2, and just start cutting and trying.
     
  6. Aug 21, 2016 #5
    I'm getting Sm as the unknown metal; is that right?
     
  7. Aug 21, 2016 #6

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    Samarium? Mental arithmetic checks out.
     
  8. Aug 22, 2016 #7
    Awesome. What happens to the hydrogen though?
     
  9. Aug 22, 2016 #8

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    It "fizzes" away up the chimney/hood/exhaust system.
     
  10. Aug 22, 2016 #9
    I mean why is it not represented in the chemical equation
     
  11. Aug 22, 2016 #10

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    "M" + nHCl = (n/2)H2 + "M"Cln; it is.
     
  12. Aug 23, 2016 #11

    Borek

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    Hydrogen is present in the reaction equation and we can even easily calculate its amount. We just ignore it as it is not necessary for solving the problem (just like the amount of HCl doesn't matter, as long as there is enough of it).

    The only part of the equation that matters is

    Sm → SmCl3

    It preserves the only important information here - stoichiometric ratio between Sm and SmCl3. When solving this particular problem everything else is just a noise.
     
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