March 30, 2017

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propane (c3h8) is burned in oxygen to produce carbon dioxide and water. The heat of combustion of propane is -2012 kJ/mole. How much heat is given off when 3.0 mol C3H8 are react with 10.0 mol O2?

Thanks, and please if possible show all steps!

  • Chemistry help (DrBob222) - ,

    First you must determine which is the limiting reagent.
    C3H8 + 5O2 ==> 3CO2 + 4H2O

    3 mol C3H8 x (3 mol CO2/1 mol C3H8) = 3*3/1 = 9 mol CO2

    10 mol O2 x (3 mol CO2/5 mol O2) = 10*3/5 = 6 mol CO2
    So the limiting reagent is O2. You are reacting 5 mol O2 to produce 6 mols CO2.
    You get 2012 kJ/5*32 g oxygen. Therefore, for 320 g (10*32 g O2) you will get
    (2012 kJ/160) x 320 = ?

    You could do it another way with C3H8 but you must determine how much of the C3H8 reacted since some of it is left. 10 mols O2 x (1 mol C3H8/5 mol O2) = 2 mol C3H8 reacted.
    Then (2012 kJ/44 g C3H8) x (2*44 g C3H8) = ?
    Hope this helps. Next time tell me what your trouble is so I can concentrate on that. This way I'm just doing your work for you and you're trying to figure out where you went wrong.

  • Chemistry help (DrBob222) - ,

    ok will do, I just honestly had no idea how to do this. Can you explain how you got "2012 kJ/5*32 g oxygen. Therefore, for 320 g (10*32 g O2) you will get
    (2012 kJ/160) x 320 = ? " this part?

    thanks :)I don't understand where the grams came from, like I know it is from O2, but why can you not use moles? why do you have to use grams?

  • Chemistry help (DrBob222) - ,

    You don't have to use grams and you CAN use mols. I don't and the reason I don't use mols is it keeps me from making a mistake.
    The equation tells you that you get 2012 kJ/5 mols O2 or 2012 kJ/1 mol C3H8. (and this is where I make my mistake---I am want to reason that since it is 2012 kJ/MOL that must mean 2012 kJ/mol C3H8 OR 2012 kJ/mol O2 and it isn't. It's 2012 kJ/1 mol C3H8 or 2012 kJ/5 mol O2. By converting to grams I don't make the error).
    5 mols O2 is 5*32 = 160 g O2.
    1 mol C3H8 is 44 g.
    So you get 2012 kJ/160 g O2 or you get 2012 kJ/44 g C3H8. Same statement.
    You want to know, for 10 mols O2, how much energy is released so that is
    (2012 kJ/160g O2) x 320 g O2 = ?

    It may be a little less confusing to you if we don't convert to grams (the problem is in mols anyway)
    So you used 10.0 mols O2 and 2 mols C3H8.
    2012 kJ/1 mol C3H8 x 2 mol C3H8 used = 2012*2 = ?
    2012 kJ/5 mol O2 x 10.0 mol O2 used = 2012*2 = ?

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