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Acetanilide produces one molecule of hydroge bromide

  1. Feb 2, 2005 #1
    If each molecule of acetanilide produces one molecule of hydroge bromide, what volume of HBr gas should be evolved from the reaction?

    So to solve this, ONE molecule of acetanlide will produce ONE molecule of hydrogen bromide...

    then, the molecular weight of hydrogen bromide produced will have a mass of 82 grams...

    Ok, am i doing this problem correctly so far?

    So what step do i have to go to next, i am confused on how to approach this problem, thanks
  2. jcsd
  3. Feb 3, 2005 #2


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    You must know the starting amount of acetanilide, that is, the mole amount. 1:1 ratio will give that moles of HBr after the reaction is completed.

    Note that the molecular weight is 82 grams/mol, so just one molecule is only 82 amu (atomic mass unit). Don't use grams when dealing with atoms, molecules, etc, without multiplying them with Avogadro's number.
  4. Feb 3, 2005 #3
    Since you want to know the volume, you need not even care about the molecular or molar weight of HBr, as you can relate the amount of gas molecules directly to the volume they demand. At room temperature the molar volume is 24,0 dm3/mol I think, or was it 22,4 at room temperature and 24,0 at some other? PV=nRT (OK actually only for ideal gases but it normally holds fairly well in order to use it for real gases) so V_molar = V/n = R*T/P = 8,31 J/K * 298 K (oh, 25 deg celsius is warm to me but I think it is called room temperature) / 1 atm (and convert atm to N/m2 = Nm/m3 = J/m3).

    OK, say you have 1 g of acetanilide (dunno the substance), and just hypothetically (but totally non-physically) it has molar mass 2 g/mol (yeah, as I said non-physical, it should of course be more, but I just pick a figure that is ez to count with).
    1 g / (2 g/mol) = 0,5 mol, which (if the "ONE to ONE" assumption is correct) is the amount of HBr produced.
    0,5 moles of HBr use 0,5 mol * 24 dm3/mol = 12 dm3 = 12 Litres

    Is this the answer to your question? No, I guess you have some more reasonable numbers to put into the equations. And moreover, as chem_rt noted, you didn't state the starting quantity, nor the type of quantity to start with (volume? mass? it almost sounds like part of a larger problem where you should have calculated some quantity already)

    Hope this helps!
    Keep up the hard work :wink:
    Last edited: Feb 3, 2005
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