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Need a little push in the right direction

  1. Apr 14, 2007 #1
    I have two questions.


    a coffe-cup clorimeter, 125 mL of a 2.75 M solution of silver nitrate at 25.00 degrees C is comined with 350 mL of a 4.00 M strontium chloride solution, aslo at 25.00 C. The temperature of the final solution is 32.15 degrees C. If the density of the final solution is 1.25 g/mL and the heat cpacity of the final solution is 4.33 J/C*g, calculate delta H in kJ/mol of silver chloride formed.

    I'm not sure if I'm answering this correctly. Heres what I did:

    I found that there is .34375 moles of AgNO3 and 1.4 moles of SrCl2

    Using density = M/V, I found the mass to be 608g

    I then calculated the heat absorbed to be 18.8 KJ

    With this, I determined that the amount of moles burned = 349

    I assumed a constant pressure, so that delta H = delta E = 18.8/349, thus ending up with .0539 kJ/mol

    Does this seem right at all?


    I have no idea how to do this problem, or even attempt to do it:

    I'm asked to calculate the enthalpy change for CaC2 + 2 H2O = Ca(OH)2 + C2H2, given:
    Ca + 2C = CaC2 delta H =-62.8 kJ
    Ca+(1/2)O2 =CaO delta H =-635.5 kJ
    CaO + H2) = Ca(OH)2 delta H =-653.1kJ
    C2H2 + (5/2)O2 = 2 CO2 + H2O delta H = -1300kJ
    C + O2 = CO2 delta H =-393.51kJ

    Can anyone help me out here? I don't know anyone who has taken a chemistry class and my book isn't that good (I got the cheapest one I could find in english). An explination or any good websites discussing problems such as these would be highly appreciated.

  2. jcsd
  3. Apr 15, 2007 #2
    Code (Text):

    Okay, the idea here is manipulate the given equations
    into the form you want, noting that you may
    switch the direction of the equations provided you
    change the sign on delta H, and you can adjust
    the stochiotic (sp?) values provided you multiply
    or divide delta H by that same amount. Here goes:

    Want: CaC2 + 2 H2O = Ca(OH)2 + C2H2


    (1) Ca + 2C = CaC2 delta H =-62.8 kJ
    (2) Ca+(1/2)O2 =CaO delta H =-635.5 kJ
    (3) CaO + H2O = Ca(OH)2 delta H =-653.1kJ (<-- CHANGED to H2O)
    (4) C2H2 + (5/2)O2 = 2 CO2 + H2O delta H = -1300kJ
    (5) C + O2 = CO2 delta H =-393.51kJ

    CaC2 = Ca + 2C                 delta H = 62.8 kj (using (1))
    CaO + H20 = Ca(OH)2            delta H = -653.1 kj (using (3))
    2 CO2 + H20 = C2H2 + (5/2)O2   delta H = 1300 kj (using (4))

    Lets add these together and see what we get and work from

    CaC2 + CaO + 2 H2O + 2 CO2 = Ca + 2C + Ca(OH)2 + C2H2 + (5/2)O2

                                   delta H = 62.8 + -653.1 + 1300
                                           = 709.7 kj

    so next you may want to get rid of CaO on the LHS, so you might

       Ca + (1/2) O2 = CaO         delta H = -635.5 kj

    so we have:

    CaC2 + CaO + 2 H2O + 2 CO2 + Ca + (1/2) O2 =

    Ca + 2C + Ca(OH)2 + C2H2 + (5/2)O2    delta H = 709.7 - 635.5 kj
                                                  = 74.2

    But some of the elements cancel (Ca on the left and
    Ca on the right, etc), so we have

    CaC2 + CaO + 2 H2O + 2 CO2 = 2C + Ca(OH)2 + C2H2 + 2 O2

                                          delta H = 74.2

    Note what happened to O2, we had (1/2) on the left and
    (5/2) on the right, (5/2) - (1/2) = 2.

    Anyway you sort of work it like that until you get
    the right equation.
  4. Apr 15, 2007 #3
    Okay, that makes sense. Thanks a lot!

    How does the first answer look? I feel like I did it wrong. Is it safe to assume constant pressure, or should I be trying to solve this some other way?

    Thanks again for your help.
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