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Homework Help: Reaction of strontium nitrate + copper (II) sulfate pentachloride?

  1. Jan 17, 2009 #1
    1. The problem statement, all variables and given/known data
    We are doing a lab where we react 2.00 g of strontium nitrate (dissolved in water) + 2.00 g of copper (II) sulfate pentahydrate (dissolved in water). I need to make sure that I did the balanced chemical equation right?
    we eventually have to find the theoretical yield.

    3. The attempt at a solution
    Sr(NO3)2 (aq) + CuSO4(aq) -> SrSO4(s) + CU(NO3)2 (aq)

    I'm not sure if i put the pentahydrate (5H2O) into the equation because i'm not sure where it goes, but i think i have to because the 2.00 g of copper (II) sulfate pentahydrate is partially H20?
    Last edited: Jan 18, 2009
  2. jcsd
  3. Jan 18, 2009 #2


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    Staff: Mentor

    Seems like you are on the right track, although I am not sure if I have ever heard about copper (II) sulfate pentachloride :wink:

    When calculating moles of copper sulfate you have to incorporate pentahydrate part into molar mass.

    Try to write net ionic reaction, you will get rid of the copper then.

    Copper is a little bit tricky here, as in the solution it is not present in the form of pentaaqua complex, but rather hexaaqua, which means that it reacts with water when dissolved. To make things more complicated copper nitrate solid will be hydrated as well, but for details you should consult some handbook. Net ionic reaction allows you to ignore all these problems.
  4. Jan 18, 2009 #3
    oops i meant pentahydrate, haha sorry, i edited it. :)

    will it be just ok if i use this equation, since that looks really complicated and I'm not really dealing with the other stuff, just the precipitate SrSO4

    Sr(NO3)2 (aq) + CuSO4.5(H2O) (aq) -> SrSO4(s) + Cu(NO3)2 (aq) + 5(H2O)

    then i would
    1. find the mol for Sr(NO3)2 and CuSO4.5(H2O)-> 0.009450 mol and 0.008009 mol
    2. find the limiting reagent-> CuSO4.5(H2O)
    3.use ratios to find the theoretical yield for SrSO4 -> 1.471 g
    4. percentage yield, i got 1.41 g in my experiment, so -> 95.9%

    would this be right?
  5. Jan 18, 2009 #4


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    Staff: Mentor

  6. Jan 18, 2009 #5
    alright, thanks for the help !
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