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Homework Help: Need some tips about solving confusing Stoic Problem

  1. Nov 9, 2017 #1
    1. The problem statement, all variables and given/known data
    What mass of iron(3) chloride should be added to 450 ml of 0.132 M CaCl2 to give a solution with final chloride concentration of 0.555M

    There are a lot of numbers, and I get confused on how to convert it to get mass of iron(3) chloride!!
    2. Relevant equations
    M = n/L

    3. The attempt at a solution
    ? g iron(3) chloride (162.2 g/mol)
    .450 L of 0.132 M CaCl2 gives 0.0594 mols
    0.555 M Cl- = 0.555 mols/L

    I know the mole to mole ratio we would use is 2:3 but like, i get stuck at trying to start with the right numbers in my dimensional analyst...

    my attempt:

    0.555 mols/ L Cl- * (0.45 L CaCl2) * (3 mols Cl/2molsCl) *(1mol FeCl3/3molsCl ) = 0.12 (which is wrong).
     
  2. jcsd
  3. Nov 10, 2017 #2
    Firstly, you must find how many mols of Cl- you must add to the existing solution to achieve chloride concentration of 0,555 M.

    Then, you must think how many moles of FeCl3 are needed to provide these mols.

    Finally, using the molar mass you can calculate the respective mass of FeCl3.
     
  4. Nov 11, 2017 #3

    epenguin

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    Even before that you'd have to know what is the molarity of chloride in the existing CaCl2 solution?
     
  5. Nov 11, 2017 #4
    Of course,

    CaCl2 disassociates by the reaction:

    CaCl2 ## \to ## Ca2+ + 2Cl-.

    So, the molarity of chloride is two times the molarity of CaCl2. Right?
     
  6. Nov 11, 2017 #5

    epenguin

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    Right enough - so give it a number.
     
  7. Nov 11, 2017 #6
    I figured out that there are 0.1188 mols of Cl- in the solution but how do I know how many moles Cl to achieve .555 M cl?
     
  8. Nov 11, 2017 #7
    Hi. I just started the problem again and I'm still confused. I got the moles of Cl that's in the current solution but I'm not sure how to find how many more moles we need to get the .555 M Cl concentration
     
  9. Nov 11, 2017 #8

    Borek

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    Staff: Mentor

    Typically all you need in such situations is subtraction.
     
  10. Nov 11, 2017 #9
    I know that but isn't. .555 M like .555 mols per liter? How can I get mols so that I can subtract it by .1188
     
  11. Nov 11, 2017 #10

    epenguin

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    You can think in moles or you can think in molarity which is as you say moles per litre, and usually you will have to go a bit back and forth between them.

    There is not just one right way. However I would find just a bit easier since you are given a molarity to start with, and you aim to finish with another molarity, to answer my question with molarity of chloride 0.132 × 2 = 0.264 M. You are asked to achieve final molarity of 0.555 M - so what's the molarity in chloride you need to add - see Borek, and it does that make sense?
     
  12. Nov 11, 2017 #11
    So. I should start with .555 M of chlorine ?
     
  13. Nov 11, 2017 #12
    I'm am still lost ;(. Here's what I got

    Untitled-1.jpg
     
  14. Nov 11, 2017 #13

    epenguin

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    Sorry I just can't read that. In future if you must post Photos of homework pages, which is discouraged and certainly not necessary for such a simple problem as this, please use an app like the free DocScanHD to render them readable and post them the right way up.

    OK, so for a litre you want there to be 0.555 moles of chloride, you have got 0.264 moles of chloride (see #10) - how many more moles of chloride do you need there to be in that litre? How difficult is that?
     
  15. Nov 11, 2017 #14

    I like Serena

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    Hey Lori! :)

    I have trouble reading what you've got as well, and moreover, I have trouble making sense what everything is, and how they are connected.

    To solve a problem like this, it usually helps to write down all given information first, and make a before-after matrix.

    Something like:
    $$\begin{array}{ll} V=450\text{ mL}\\
    c(CaCl_2)=0.132\text{ mol/L}\\
    c_{after}(Cl^-)=0.555\text{ mol/L}\\
    M(FeCl_3)=162.2\text{ g/mol}\\
    \begin{array}{|c|c|c|}
    \hline
    &Before & After & Delta \\
    \hline
    c(Cl^-) & ? & 0.555\text{ mol/L} & - \\
    n(Cl^-) & ? & ? & ?\\
    n(FeCl_3) & ? & - & -\\
    \hline
    \end{array} \\
    m(FeCl_3) =?\end{array}$$
    Relevant formulas:
    - moles are concentration times volume (##n=c\cdot V##),
    - mass is moles times molar mass (##m=n\cdot M##).

    Can we find all the question marks?
     
    Last edited: Nov 12, 2017
  16. Nov 11, 2017 #15
    I'm Sorry, but I expected that there would be confusion cause my calculation is all over the place!! I will follow your strategy. I'll let you know what I get!
     
  17. Nov 11, 2017 #16
    Ok, i got that we need 0.291 mols Cl- /liter


    Oh i see! We need .291 mols Cl- /liter, but we have 0.450 L of CaCl2, so if we multiply these numbers together , we get the mols we need of Cl-. With 0.13095 mols Cl-, calculate moles of FeCl3 we need by using ration 1 fecl3 to 3 moles cl-. Converting to grams in FeCl3 with molar mass, we get 7.08 grams???

    Is it is 7.08 grams of FeCl3???? :O


    I'm pretty sure i spent like 3 hours on this problem, but i realize that i just need to think about and ask the same questions that you guys asked. Sigh!

    Thanks for your guys patients >.<
     
    Last edited by a moderator: Nov 11, 2017
  18. Nov 12, 2017 #17

    Borek

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    7.08 g of FeCl3 it is.

    We don't ask them without reason.

    Note: in the real lab you would have serious problems finding anhydrous FeCl3, as this compound is typically sold as hexahydrate (FeCl3⋅6H2O). That means you would need almost 12 g of the solid.
     
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