Stoichiometry of Precipitation Reactions

  • #1

Homework Statement



What mass of iron(III) chlorate is needed to precipitate all the chromate ions from 234 mL of a 0.100 M lithium chromate solution.


Homework Equations





The Attempt at a Solution



we just started precipitates and my teacher didnt explain the procedure very well so if u can get me on the correct track for solving it, it would be appreciated.

i made a balanced equation for the reaction
2Fe(ClO3)3 + 3Li2CrO4-->6LiClO3 + Fe2(CrO4)3

then the net ionic equation 2Fe3+ + 3CrO42--->Fe2(CrO4)3

then i used molarity equations to find 0.1M=xmol/.234L --> 0.0234 mol Li2CrO4

but i am not sure where to go from here, any help?
 

Answers and Replies

  • #2
symbolipoint
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You have the correct number of moles of lithium chromate. What is the ratio of Iron(III) Chlorate to Lithium Chromate? How many moles then of the Iron Chlorate? So find the formula weight of Iron Chlorate and compute this mass as grams.
 
  • #3
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Use the mole ratio to go from mols of lithium chromate to mols of iron chlorate. Once you have mols of iron chlorate, calculate the formula mass and just multiply.

Mols of Lithium Chromate * 2/3 = Mols of Iron Chlorate
(mole ratio)

Mols of Iron Chlorate * Formula mass of Iron Chlorate = Answer!
 
  • #4
Great thanks guys, I got it now. My teacher made it pretty confusing. I appreciate the help.
 
  • #5
symbolipoint
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Maybe maybe not about your teacher. Some of them can be like that, not most. You find your mole ratios among reactants or products from the written reaction. Your mole ratios are essentially conversion factors.
 
  • #6
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Maybe maybe not about your teacher. Some of them can be like that, not most. You find your mole ratios among reactants or products from the written reaction. Your mole ratios are essentially conversion factors.
Just to emphasize, I've always found it best to consider mole ratios as conversion factors. I know some teachers don't really push that through your head, but I find it much easier to think of them exactly that way.
 
  • #7
Yeah when you think about them that way it makes it a lot easier. Once I started to do a lot of these problems it becomes pretty easy
 

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