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Calculate the volumes of air and hydrogen gas

  1. Sep 22, 2004 #1
    If anyone knows how to do this problem, could you please give me some
    hints?? I don’t need the answer just some help as to where to start....

    Metallic molybdenum can be produced from the mineral molybdenite, MoS2.
    The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide.
    Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen
    gas. The balanced equations are:

    MoS2 (g) + 7/2 O2 (g) -------> MoO3 (s) + 2SO2 (g)

    MoO3 (s) + 3H2 (g) ------------> Mo (s) + 3H2O (l)

    Calculate the volumes of air and hydrogen gas at 17 deg. C and 1.00 atm
    that are neccessary to produce 1.00 x 10^3 kg of pure molybdenum from
    MoS2 . Assume air contains 21% oxygen by volume and assume 100% yield
    for each reaction.
  2. jcsd
  3. Sep 23, 2004 #2


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    Gold Member


    1) One ton of pure molybdenum is not a useful definition, it's better to convert it into moles.

    2) As one mole of molybdenum is produced by three moles of hydrogen, you can find how many moles of H2 is needed, and then convert it into volume by [tex]P*V=n*R*T[/tex].

    3) The mole of molybdenum is the same as molybdenum(VI)oxide; just put this number in the first reaction, and multiply it by 3,5 to obtain how many moles of pure oxygen is needed. Then convert it into volumes.

    4) If 100% of oxygen is present in the environment, the volume you find will be enough, but only 21% of oxygen is present in the air. You'll have to divide something to learn the actual volume.

    Hope these help.

  4. Sep 23, 2004 #3
    Thank you for your help chem_tr, for this problem as well as the one on effusion rates! Helped a lot. :)
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