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Stuck on Reaction Enthalpy Question

  1. Nov 30, 2017 #1
    1. The problem statement, all variables and given/known data
    What mass of methane must be combusted with excess oxygen to generate enough heat to warm 675 g of water by 25.0 C? (assume no heat lost to surroundings)

    2. Relevant equations
    Specific heat of H20= 4.184 J/g*C
    q = mCT
    balanced equation: CH4 + 2O2 -> CO2 + 2H2O
    3. The attempt at a solution
    I'm kinda stuck here. Here's what i found...

    from stoic, found that 675g H20 is 37.5 mols of H20 and that its 18.75 mols of CH4

    I also figured out that the energy the water absorbs from the reaction is 70605 Joules from m*C*T

    every reaction of 18.75 mols of CH4 creates 16693.125 KJ but we need 70605 Joules. Thus, 70605 Joules - 16693.125 KJ = 53911.875 J needed ? I kinda paused here though cause im not sure if it's right!
  2. jcsd
  3. Nov 30, 2017 #2

    Charles Link

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    You are warming 675 g of water (in a pot) with a temperature increase ## \Delta T=25^o \, C ##. You don't need to create 675 g of water. (The water is present in the pot before the reaction). You should know the heat capacity of water is 1.00 cal/(gram-degree centigrade). This allows you to compute the ## \Delta Q ## that you need. What is ## \Delta H ## per mole for this methane reaction? Then you can compute the number of moles of methane you need.
  4. Nov 30, 2017 #3
    Is it right to say that the Q of water (m*C*T) is the heat we need to be generated by the combustion?
  5. Nov 30, 2017 #4

    Charles Link

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    Yes .## Q=mC \, \Delta T ## and ## C=1.00 cal/(gram-degree Centigrade) ##.
  6. Nov 30, 2017 #5
    How come we use C = 1.00 instead of 4.184?
  7. Nov 30, 2017 #6
    Oh ! 4.184 J = 1 cal

    Ok.. so i found that Q for water is 70.605 Joules and that 1mole CH4/890.3 KJ, using dimensional analysis, i will need 0.079 moles of CH4 which is 1.27 g CH4 needed
  8. Nov 30, 2017 #7

    Charles Link

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    Editing... (you added something)... I think Q should be 70,605 Joules. (or 70.605 kJ). Otherwise, what you have looks correct. :)
  9. Nov 30, 2017 #8
    oops! you're right. I meant kilojoules :p
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